# Array Manipulation

# Array Manipulation

amansbhandari + 248 comments Guys, below is the code in O(n) time complexity and O(1) Auxiliary space

`#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { long int N,K,p,q,sum,i,j,max=0,x=0; cin>>N>>K; long int *a=new long int[N+1](); for(i=0;i<K;i++) { cin>>p>>q>>sum; a[p]+=sum; if((q+1)<=N) a[q+1]-=sum; } for(i=1;i<=N;i++) { x=x+a[i]; if(max<x) max=x; } cout<<max; return 0; }`

amitkumarjnv2 + 10 comments Your approach is brilliant. Hats off

yugesh + 8 comments can you please explain the logic behind that.?

amitkumarjnv2 + 10 comments see, you are adding sum to a[p] and adding negative sum at a[q+1]. which make sure that when you add element from a[p] to a[q] sum is added only once and it should be subtracted at a[q+1] as this sum span from p to q only. Rest array element are either 0 or some other input sum. max of addition will be output. refer to above code for p, q, and sum.

amansbhandari + 0 comments [deleted]lancerchao + 12 comments Instead of storing the actual values in the array, you store the difference between the current element and the previous element. So you add sum to a[p] showing that a[p] is greater than its previous element by sum. You subtract sum from a[q+1] to show that a[q+1] is less than a[q] by sum (since a[q] was the last element that was added to sum). By the end of all this, you have an array that shows the difference between every successive element. By adding all the positive differences, you get the value of the maximum element

manjiri + 3 comments Important points to note in this solution.

1)the first element of array a will always remain zero since 1<= a <=b <=n; 2

2)the second element of array a is the second element of the array after m operations.

vineet_ahirkar + 27 comments Same solution translated in python -

`n, inputs = [int(n) for n in input().split(" ")] list = [0]*(n+1) for _ in range(inputs): x, y, incr = [int(n) for n in input().split(" ")] list[x-1] += incr if((y)<=len(list)): list[y] -= incr; max = x = 0 for i in list: x=x+i; if(max<x): max=x; print(max)`

jbwhite + 3 comments And the same approach in ruby -- I claim no credit for working this out -- I wrote this after reading the comments and code posted here.

Just thought I'd add a ruby solution for anyone looking for one.

N, M = gets.chomp.split(' ').map(&:to_i) # create array of zeros of length N + 1 arr = Array.new(N + 1, 0) M.times do # cycle through and get the inputs start, finish, value = gets.chomp.split(' ').map(&:to_i) # increment value at start of sequence arr[start - 1] += value # decrement value at first position after sequence arr[finish] -= value end tmp = 0 max = 0 arr.each do |value| # step through summing array tmp += value # capture the max value of tmp max = tmp if max < tmp end puts max

kaustubh_p16 + 11 comments Same solution in C

int main() { long long int n,k,i,max=0,x=0; scanf("%lld %lld",&n,&k); int *a=(int *)malloc(sizeof(int)*(n+1)); for(i=0;i<n;i++){ *(a+i)=0; } for(i=0;i<k;i++){ long long int c,d,g; scanf("%lld %lld %lld",&c,&d,&g); *(a+c)+=g; if(d+1<=n){ *(a+d+1)-=g; } } for(i=1;i<=n;i++){ x+=*(a+i); if(max<x){ max=x; } } printf("%lld",max); /* Enter your code here. Read input from STDIN. Print output to STDOUT */ return 0; }

acdevbox1 + 9 comments Same solution in C#

using System; using System.Collections.Generic; using System.IO; class Solution { static void Main(String[] args) { string[] inString = Console.ReadLine().Split(' '); uint[] initParams = Array.ConvertAll(inString, UInt32.Parse); uint n = initParams[0]; uint m = initParams[1]; long[] numList = new long[n+1]; for(int i=0; i<m; i++) { string[] opString = Console.ReadLine().Split(' '); uint a = UInt32.Parse(opString[0]); uint b = UInt32.Parse(opString[1]); long k = long.Parse(opString[2]); numList[a] += k; if(b+1 <= n) numList[b+1] -= k; } long tempMax = 0; long max = 0; for(int i=1; i<=n; i++) { tempMax += numList[i]; if(tempMax > max) max = tempMax; } Console.WriteLine(max.ToString()); } }

jmgallagher + 22 comments Same solution in Java

Scanner scan = new Scanner(System.in); int n = scan.nextInt(); int m = scan.nextInt(); //This will be the "difference array". The entry arr[i]=k indicates that arr[i] is exactly k units larger than arr[i-1] long[] arr = new long[n]; int lower; int upper; long sum; for(int i=0;i<n;i++) arr[i]=0; for(int i=0;i<m;i++){ lower=scan.nextInt(); upper=scan.nextInt(); sum=scan.nextInt(); arr[lower-1]+=sum; if(upper<n) arr[upper]-=sum; } long max=0; long temp=0; for(int i=0;i<n;i++){ temp += arr[i]; if(temp> max) max=temp; } System.out.println(max);

Sarthak10 + 1 comment [deleted]jmgallagher + 1 comment It is the same logic as mentioned above.

archanapradeep + 3 comments Hi i dont understand how the difference array works. What is the logic behind adding at one index and subtracting at the other and taking its sum?

boris93 + 3 comments You can try to visualize the array as steps / stairs

We are just noting down the bump ups and bump downs

chikmid + 2 comments I still havent understood this logic.Even though i implemented this logic in java with ease,i dont understand how this logic helps us arrive at the solution.

jricaldi + 2 comments me netheir, I am looking for the maths here, I am pretty sure the solution has a math method. Somebody here wrote "Prefix sum".

mvanwoerkom + 0 comments I tried an answer in the spirit of digital signal processing here.

kemal_caymaz + 10 comments After thinking like that i also understood the logic the solution.

Let's think our summing part input like that {A B S} = {1 3 100} {2 5 150} {3 4 110} {2 4 160}

Instead of writing all elements of array we can write maximum value at just starting and ending indexes to have less writing operation. So, after first input row, array can be something like that.

0 100 0 100 0 0 0 0 0

But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.

Instead of doing like that we can write S value to index A and -S value to B+1, so it is still similar logic. Starting from A to B all indexes have S value and rest of them have less than these indexes as S as. Now the array is like that:

0 100 0 0 -100 0 0 0 0

While calculating second row, we are writing 150 to index 2 and -150 to index 6. It will be like that: 0 100 150 0 -100 0 -150 0 0

If we write array with old method, which means that all numbers calculated one, it will be: 0 100 250 250 150 150 0 0 0

It shows that value of index 2 is : 100+150 = 250. Value of index 5: 100 + 150 + (-100) = 150. So by calculating with the solution written above, instead of writing all numbers, we are writing changes at edge indexes.

krishnna + 1 comment check it out here, you will get all your doubts solved https://www.geeksforgeeks.org/difference-array-range-update-query-o1/

tejaspatil + 1 comment Below link will also help to understand theory behind it. https://www.geeksforgeeks.org/constant-time-range-add-operation-array/

codam + 9 comments Same solution in Javascript

var arr = []; var max = 0; // init each element of arr to 0 for (let l = 0; l < n; l++) { arr[l] = 0; } // for each sum operation in queries for (let i = 0; i < queries.length; i++) { // update arr with number to add at index=queries[i][0] and number to remove at index=queries[i][0]+1 => this will allow us to build each element of the final array by summing all elements before it. The aim of this trick is to lower time complexity arr[queries[i][0]-1] += queries[i][2]; if (queries[i][1] < arr.length) { arr[queries[i][1]] -= queries[i][2]; } } for (let j = 1; j < n; j++) { arr[j] += arr[j-1]; } for (let k = 0; k < arr.length; k++) { max = Math.max(max, arr[k]); } //max = Math.max(...arr); // not working for big arrays return max;

kamnijais + 1 comment Hey, I did the code in Java8 and my code is getting failed for input type - where only single value is present in a row of array. meaning only left index value is provided and right and k value is missing from array. So can you help me how to solve this issue?

h1017691668 + 0 comments you could post your code,and we can check it out

mevtew + 3 comments simpler in es6:

`function arrayManipulation(n, queries) { let arr = new Array(2*n).fill(0); let max = 0;`

`queries.forEach((item) => { arr[item[0]] += item[2]; arr[item[1] + 1] -= item[2]; }); arr.reduce((prev, curr, idx) => { const sum = prev + curr; if (sum > max) { max = sum; } return sum; }) return max;`

`}`

Rockyfive + 1 comment I did something pretty similar, just with a little bit more readable forEach:

`const arr = new Array(n).fill(0); let result = 0; queries.forEach(([a, b, k]) => { arr[a - 1] += k; if (b < arr.length) { arr[b] -= k; } }); arr.reduce((a, b) => { const acc = a + b; result = Math.max(result, acc); return acc; }, 0); return result;`

susahin80 + 3 comments This produces wrong answer in some of the tests.

Kanahaiya + 2 comments Hi,

try this. Here is the video tutorial for my solution O(n+m) complexity.

https://www.youtube.com/watch?v=hDhf04AJIRs&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA

Would really appreciate your feedback like and comment etc. on my video.

Fuchsia + 1 comment It is a good video. I understood the algorithm clearly.

Kanahaiya + 1 comment thanks @chandraprabha90.

but it would be great if you can please provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.

Grozny + 2 comments Could you add subtitles? I tried watching it but couldn't quite understand your accent through the audio

Kanahaiya + 0 comments Hi Grozny,

I appologize for my bad sound quality but i am trying to improve it. but it will be very difficult to add subtitle for this video because its around half an hour which explains all the concepts in deep.

Making this long video took lot of effort and time now adding a subtitle will be very tedious without any support.

Will suggest you to try automatic transcribe feature from youtube to translate it.

Anyway thanks for watching.

Kanahaiya + 0 comments Hi Grozny,

I have added subtitle for this tutorial and I hope it will you to understand logic with more clarity.

willsandalls + 0 comments [deleted]willsandalls + 0 comments I had the same issue. my mistake was decrementing lower and upper. you don't decrement upper, the difference array needs to show it went down AFTER then last index, not within.

michael_yiming_1 + 0 comments [deleted]yozaam + 0 comments Thought to myself ....no js solutions here? then I find 3 of them, Js, ES6, readable wow.

jpeavey451 + 0 comments Your last for loop isn't needed. You can move Math.max to the previous for loop.

lyons_mr + 0 comments [deleted]tyler_e_austin + 0 comments [deleted]tyler_e_austin + 0 comments Added some ES6 syntax suger...

const arrayManipulation2 = (n, queries) => { const arr = new Array(n).fill(0); let max = 0; for (let i = queries.length - 1; i >= 0; i--) { const [a, b, k] = queries[i]; arr[a - 1] += k; if (b < arr.length) { arr[b] -= k; } } for (let j = 1; j < n; j++) { arr[j] += arr[j - 1]; } for (let k = arr.length - 1; k >= 0; k--) { max = Math.max(max, arr[k]); } return max; };

shubhanksrivast1 + 0 comments Got stuck because of this

max = Math.max(...arr); // not working for big arrays

Thanks man!

brandon53 + 0 comments You could simplify your code a smidge, and save a little processing power, by removing your final for loop, and putting in an if check in the second to last loop, like this:

for (let j = 1; j < n; j++) { arr[j] += arr[j-1]; if (arr[i] > biggest) { biggest = arr[i]; } }

rvshah1992 + 1 comment [deleted]rvshah1992 + 0 comments Perfect! Could you please explain me the thought process behind the solution?

dawar_dhawal + 0 comments Thank you @Kemal_caymaz for the explanation, I have found this very useful.

VE5T1GE + 0 comments Thank you!

prashant_kumar_3 + 0 comments thnx

rohanmehta0077 + 1 comment super awesome X 1000!!!

can you expalin this:

**But the problem is here that even we didn't write anything, value of index 2 is also 100. When we wanted to continue with second step we have to check whether index 2 is between indexes of first row operation or not.**Kanahaiya + 2 comments Hi,

I have created a video tutorial for you and uploaded the same on youtube. Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

marwan_daar + 1 comment fantastic video, thank you!

Kanahaiya + 0 comments most welcome.

but if you dont mind can you please leave the same comment along with like, dislike on my video. it motivates me and help others too find the solution over internet.

naimulhaquesagar + 0 comments impressive ...

bhavyseth + 0 comments thanks bro..

Kanahaiya + 0 comments Hi,

Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

manshujain2020 + 1 comment Thanks a lot budy for your fantastic explanation ! Your thinking is really amazing like you . Good job.

Kanahaiya + 0 comments most welcome. It would be great, if you can provide your feedback like, dislike , comment etc. on my video. It motivate me to do more for you all

ashishjjha03 + 1 comment hey used this logic its failing for 10 test cases with large inputs

Kanahaiya + 0 comments Hi,

Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

amitojsingh_edc1 + 0 comments Thanks, your explanation is very helpful

Krishnasaivamsi + 1 comment did you got it?

shubhojeetpaul91 + 0 comments getting it correct for few cases but when both indexes are same then its giving a error message

lyons_mr + 0 comments To explain further for people confused:

We are creating a "difference array" Which shows how many steps up or down have occurred (the difference between 0 and result of each operation) and where in the array they have occurred. This way, you can see how high the max ends up and return that for the solution.

I found this explanation helpful to have this fact dawn on me after much noodling: https://www.geeksforgeeks.org/difference-array-range-update-query-o1/

rui_zhang_cmlab + 0 comments Like piling up blocks? Adding a number -> going up one step and subtracting -> down . Finally, we count how high we can go.

scirelli + 2 comments I'm still trying to figure it out myself. But if you graph result after doing the operations, you would see some rise and fall in the graph.

It looks like his solution tracks the differences between each data point. It went up by x, down by y, remained the same...etc. And his solutions finds the highest increase.

Example: 5 3

1 2 100

2 5 100

3 4 100After doing the operations you get [100, 200, 200, 200, 100] His solutions final array is [0, 100, 100, 0, 0, -100] Meaning starting at 0 the graph went up by 100, went up by 100 again, remained the same, then went back down by 100.

So the highest point is 200, the solution.

jie_lyu + 0 comments you add up all the numbers > 0 in the final list, which is 100 + 100 = 200

pillmannr1 + 1 comment One insight that might help is that we're keeping track of the change in values rather than the values themselves at each index. Therefore, by adding the k at a and subtracting it after b, we are saying that at a the total value increases by k compared to the previous element, and after b the total value decreases by k compared to b. Hope that helps!

adriana_elizond1 + 0 comments Here's the same solution in swift in case anyone needs it :).

func arrayManipulation(n: Int, queries: [[Int]]) -> Int { var sums = Int: Int for query in queries { sums[query[0]] = (sums[query[0]] ?? 0) + query[2] sums[query[1] + 1] = (sums[query[1] + 1] ?? 0) - query[2] }

`var currentmax = 0 var sum = 0 sums.sorted{ `$0.0 < $`1.0 }. compactMap{ `$0.1; sum += $`0.1; currentmax = sum > currentmax ? sum : currentmax} return currentmax`

}

coolcoder001 + 1 comment Hi , I have a doubt.

Here the indices are starting from 1. So, we should be subtracting 1 from both lower index and upper index. Here you have done so for lower index, but haven't done for upper index.

Can you please explain the reason behind this ?

xavier630 + 0 comments It's because it doesn't go back down until the element after the section ends.

eg: n = 4, a = 1, b = 2 k = 3. So we have 3 3 0 0 after reading in that line. In his array he represents this as 3 0 -3 0 ie the subtraction is the element after the last element in the section.

The reason the lower value has a "-1" is because java uses 0-indexed arrays, ie they start at 0. But the input for this question only starts at 1. So he puts the values one index lower in the array. The upper value has no "-1" for the reason in the above paragraph about subtracting after the last element in the section

sergioescala + 0 comments You don't have to do this:

for(int i=0;i<n;i++) arr[i]=0;

because

**long**by default is 0harshithakamma + 2 comments could you please explain me the working of the code?

JStardust + 0 comments i think test code bigger than long int,so we need a larger data structure

Kanahaiya + 1 comment Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

wasitshafi + 0 comments tnks @kanahaiya

lavrikilya + 0 comments [deleted]apivovarov + 0 comments no need to init array elements to 0 in Java

sbouzel + 1 comment Hi, your solution works but I am not convinced!

I don't see how do you increment all elements between lower and upper!

- arr[lower-1]+=sum; This only increments the lower element by sum.
- if(upper
- arr[upper]-=sum; I don't understand why are you substructing sum!

Can you please explain? Thanks.

jasonliuxz + 1 comment It's a difference array. He is storing the difference/the changes that should be made in each index and then runs a loop to add up all these changes. You increment the lower bound because everything inbetween the lower and upper bound should be incremented but you dont want this change to continue for the rest of the array so you decrement at (upper+1).

harshithakamma + 0 comments but how can we know about the upper index of a particular increment,when we are adding all at a once in a loop?

juank_luna + 1 comment Using Java 8 syntaxis:

public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int m = in.nextInt(); long[] output = new long[n]; IntStream.range(0,m).forEach(i -> { int a = in.nextInt()-1; int b = in.nextInt(); int k = in.nextInt(); output[a] += k; if(b < n) output[b] -= k; }); AtomicLong sum = new AtomicLong(0); System.out.print(LongStream.of(output).map(sum::addAndGet) .max().getAsLong()); in.close(); }

AlexisVaBel + 2 comments large amount of lines will crash your solution. And map fucntion need to make Boxing methinks and this takes much time

Good_Imagination + 0 comments WTF!!! It's true! I had a time out problem with test case 7 up to 12 I think and my code is good enough and didn't know what to do...until I read your comment, I removed all the empty spaces and the debugging prints (System.out.println for Java 8) and it worked :O LOL thanks.

vincent_messeng1 + 1 comment This is true, I spent too much time trying figure out why my solution was timing out. Deleted some whitespace and it ran fine.

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

tranhoangha94 + 0 comments why temp += arr[i];

DanYoo940 + 0 comments Array is already 0, you dont have to assign 0s to each array elements

AlexisVaBel + 0 comments scanner on large lines of input is not suitable solution, it reads too long

Sufiyan + 0 comments You don't need to loop the arr to put 0. Bydefault, it has the value zero when you initialize.

Sufiyan + 0 comments In java, we don't need to loop explicitly to assign zero to all the arr locations. When we create it, it has the zero as default values. I like your solution, thanks.

hspuri26 + 0 comments using an if statement inside the for loop will just contribute in complexity. you can replace it with Math.max function.

rosamohammadi + 2 comments Is there a reason all the solutions posted above are written inside main() and not the provided function arrayManipulation() ? Or did hackerrank just change this over the past few years for readability?

// Complete the arrayManipulation function below. static long arrayManipulation(int n, int[][] queries) { // initialize array with 0's of size n long arr[] = new long[n]; // each successive element contains the difference between itself and previous element for (int i = 0; i < queries.length; i++) { // when checking query, subtract 1 from both a and b since 0 indexed array int a = queries[i][0] - 1; int b = queries[i][1] - 1; int k = queries[i][2]; arr[a] += k; if (b+1 < n) { arr[b+1] -= k; } } // track highest val seen so far as we go long max = Long.MIN_VALUE; for (int i = 1; i < arr.length; i++) { arr[i] += arr[i-1]; max = Math.max(arr[i], max); } return max; }

srodrig0209 + 0 comments Probably just for readability.

tracylmarkland1 + 0 comments I was wondering the same thing. The instructions say to complete the manipulation method, not to rewrite the main method. I assumed that it should work without timing out if I just get the manipulation method to the point where it is efficient enough.

MaxInertia + 1 comment Same solution in Golang

func arrayManipulation(n int32, qs [][]int32) int64 { a := make([]int64, n + 1) qLen := len(qs) for i:=0; i<qLen; i++ { p := int(qs[i][0]) q := qs[i][1] sum := int64(qs[i][2]) a[p] += sum if((q + 1) <= n) { a[q + 1] -= sum } } var x, max int64 = 0, 0 for i:=1; i<=int(n); i++ { x += a[i] if max < x { max = x } } return max }

Clever solution!

shevchenko_dian1 + 1 comment Can you help me see what is wrong with my code here?

func arrayManipulation(n int32, queries [][]int32) int64 { a := make([]int64, n + 1) qLen := len(qs) for i:=0; i<qLen; i++ { p := int(qs[i][0]) q := qs[i][1] sum := int64(qs[i][2]) a[p] += sum if((q + 1) <= n) { a[q + 1] -= sum } } var x, max int64 = 0, 0 for i:=1; i<=int(n); i++ { x += a[i] if max < x { max = x } } return max }

lucash3 + 0 comments Wrong name in first line:

func arrayManipulation(n int32, queries [][]int32) int64 {

}

a2zshatakshi + 3 comments import java.io.

*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*;public class Solution {

`// Complete the arrayManipulation function below. static long arrayManipulation(int n, int[][] queries) { int[] arr = new int[n]; for(int i=0;i<n;i++){ arr[i] = 0; } int m = queries.length; for(int i=0;i<m;i++){ for(int j=queries[i][0]-1;j<queries[i][1];j++){ arr[j] += queries[i][2]; } } int max = arr[0]; for(int i=1;i<n;i++){ if(arr[i]>max){ max = arr[i]; } } return(max); } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); String[] nm = scanner.nextLine().split(" "); int n = Integer.parseInt(nm[0]); int m = Integer.parseInt(nm[1]); int[][] queries = new int[m][3]; for (int i = 0; i < m; i++) { String[] queriesRowItems = scanner.nextLine().split(" "); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for (int j = 0; j < 3; j++) { int queriesItem = Integer.parseInt(queriesRowItems[j]); queries[i][j] = queriesItem; } } long result = arrayManipulation(n, queries); bufferedWriter.write(String.valueOf(result)); bufferedWriter.newLine(); bufferedWriter.close(); scanner.close(); }`

}

What is wrong with this code. Please help.

naimulhaquesagar + 0 comments it may be showing Terminated due to timeout error because in test case 7 there is a huge number of input 10000000 100000 (like that)

you should try prefix array sum it is so simple

Kanahaiya + 2 comments Hi,

Your solution is good but time complexity of your solution is O(n*m) which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

Here is the video tutorial for my solution O(n+m) complexity.

https://www.youtube.com/watch?v=hDhf04AJIRs&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA

Would really appreciate your feedback like and comment etc. on my video.

naimulhaquesagar + 1 comment your video is helpful

Kanahaiya + 1 comment thanks. But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.

roneet_kumar_sh1 + 1 comment Your video helped me understand the algorithm. I liked the video. Thank you so much

Kanahaiya + 0 comments thanks. But if would be great ,if you can provide your comments, like, dislike on my video how i did it..It motivates me to create better content for my audience.

Mr_Mendiratta + 0 comments Thanks @Kanahaiya for sharing. Your video is very helpful. I really liked the way you explained by dry running the solution. One of the best programming tutroial ever seen. Surely, will look into other videos as well.. Thanks.

vickykumar19988 + 0 comments you will get TLE. Try different algorithm which takes less time.

mhsdawas + 0 comments Thanks for the clean code and the comment! It helped more than the entire discussion!

mariadimitrova11 + 0 comments This line is unnecessary due to the fact that the default value of long in Java is 0L and there is no need for assigning them again to 0.

`for(int i=0;i<n;i++) arr[i]=0;`

KuraudoVII + 0 comments Thanks for the enlightenment.

Here is the same solution in Swift:

func arrayManipulation(n: Int, queries: [[Int]]) -> Int { var numbers = Array(repeating: 0, count: n + 1) var ans = 0 for i in 0 ..< queries.count { var a = queries[i][0] var b1 = queries[i][1] + 1 var k = queries[i][2] numbers[a] += k if b1 <= n { numbers[b1] -= k } } for i in 1...n { numbers[i] += numbers[i - 1] if numbers[i] > ans { ans = numbers[i] } } return ans }

carlosfelipetor1 + 0 comments [deleted]codebhai + 0 comments I think

for(int i=0;i

is not needed.

izambl + 2 comments Same solution in javascript

function main() { let n_temp = readLine().split(' '); let n = parseInt(n_temp[0]); let m = parseInt(n_temp[1]); let res = []; let sum = 0; let max = 0; for (let i = 0; i<n; i++) { res[i] = 0; } for(var a0 = 0; a0 < m; a0++){ var a_temp = readLine().split(' '); var a = parseInt(a_temp[0]); var b = parseInt(a_temp[1]); var k = parseInt(a_temp[2]); res[a-1] += k; if (b<n) res[b]-=k; } for (let i=0; i<n; i++) { sum += res[i]; if (max < sum) max = sum; } console.log(max); }

Jeff_Chung123 + 4 comments very smart,here is follow output to help me understand the idea

Compilation Successful Input (stdin) 5 1 1 2 100 Your Output (stdout) Slope List [ 100, 0, -100, 0, 0 ] Actual List[100,100,0,0,0] Compilation Successful Input (stdin) 5 2 1 2 100 2 5 100 Your Output (stdout) Slope List [ 100, 100, -100, 0, 0 ] Actual List[100,200,100,100,100] Compilation Successful Input (stdin) 5 3 1 2 100 2 5 100 3 4 100 Your Output (stdout) Slope List [ 100, 100, 0, 0, -100 ] Actual List[100,200,200,200,100]

Jeff_Chung123 + 2 comments process.stdin.resume(); process.stdin.setEncoding('ascii'); var input_stdin = ""; var input_stdin_array = ""; var input_currentline = 0; process.stdin.on('data', function (data) { input_stdin += data; }); process.stdin.on('end', function () { input_stdin_array = input_stdin.split("\n"); main(); }); function readLine() { return input_stdin_array[input_currentline++]; } /////////////// ignore above this line //////////////////// function main() { var n_temp = readLine().split(' '); var listSize = parseInt(n_temp[0]); var lineNumber = parseInt(n_temp[1]); let slopList = new Array(); for(let i = 0;i < listSize;i++){ slopList.push(0); } for(var a0 = 0; a0 < lineNumber; a0++){ var a_temp = readLine().split(' '); const beginElementPos = parseInt(a_temp[0]); const endElementPos = parseInt(a_temp[1]); const addValue = parseInt(a_temp[2]); slopList[beginElementPos-1] += addValue; if (endElementPos<listSize){ slopList[endElementPos]-=addValue; } } let actualList = new Array(); let sum = 0; for (let i=0; i<listSize; i++) { sum += slopList[i]; actualList.push(sum); } let max = actualList.reduce((acc,val)=>{ return (acc>val)?acc:val; },0); console.log(max); }

ali_sid_jobs + 5 comments Guys, if you look for a clear understanding of the solution, I read a pretty clear comment down the road that clarified my mind.

Basically, when you add value from a to b you just need to know that it goes up from k and goes down of k after.

What this algo does is to register the slopes only, so we just need 2 entry, with O(1) complexity.

We just need to know that we are upping from k at the beginning and decreasing at the end.

Finally, the maximum would be...

The addition of all the slopes, that is why it's max(sum) of the tables, because the tables itself registers the slopes

mvanwoerkom + 1 comment [deleted]mvanwoerkom + 0 comments [deleted]

smaaash + 0 comments Thanks a lot!!That really helped!

agarwal_akshara9 + 2 comments Can you explain the concept of just adding add subtracting at a particular index? I mean how have we arrived to this logic?

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

Kanahaiya + 1 comment Hi,

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like and comment etc. on my video.

ozgur24 + 3 comments I didnt well understand that what will happen if b+1 is out of array bounds?

hn5624 + 0 comments it can't be out of bounds, it saids that b>n in the problem statement.

rohitshende16 + 0 comments if b+1 > n then the the addition of k from position a will continue till the last element of the array.

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

hn5624 + 1 comment Jesus christ, it all makes sense now after that graph lol, I kept wondering what drug these people were taking to arrive at this conclusion.

blunderblunder67 + 0 comments can you explain ? :)

asimparvez91 + 0 comments Very well explained (Y)

nashrahmeraj + 0 comments i have got correct output of your test cases but 3 test cases of hackerrank are getting wrong. iam not understanding whats wrong.please help me

jishangarg + 0 comments brilliant idea!

davidyc + 1 comment what about the custom input:

5 1

1 5 -100

wouldn't the output be 0? But the max would be -100 since all elements would be -100.

eellor + 0 comments There's a constraint: 0 <= k <= 10^9

In your case of negative k, the minimum value can be obtained by the similar approach. Imagine a valley rather than a mountain.

kunalwww1994 + 0 comments why are you subtracting???? "( b < n ) res [ b ] -= k;"

umutesen + 3 comments How is your method faster than the following?

using System; using System.Collections.Generic; using System.IO; using System.Linq; using System.Numerics; class Solution { static void Main(String[] args) { string[] tokens_n = Console.ReadLine().Split(' '); int n = Convert.ToInt32(tokens_n[0]); int m = Convert.ToInt32(tokens_n[1]); // Instantiate and populate an array of size N BigInteger[] arr = new BigInteger[n]; for(int i = 0; i < n; i++) arr[i] = 0; for(int a0 = 0; a0 < m; a0++){ string[] tokens_a = Console.ReadLine().Split(' '); int a = Convert.ToInt32(tokens_a[0]); int b = Convert.ToInt32(tokens_a[1]); int k = Convert.ToInt32(tokens_a[2]); // Apply operation for(int j = a-1; j < b; j++) arr[j] += k; } Console.WriteLine(arr.Max(i=>i)); } }

avmohan + 0 comments The "Apply operation part" is O(k) here. In the diff array version, apply operation is O(1)

kunalwww1994 + 1 comment hey did u passed all with this...i used same logic in C#..but passes till 6

JStardust + 0 comments your database is int? some tests data is too big

zeeshanaslamonl1 + 1 comment your code is only passing 3 test cases out of 10 . Don't post the code unless it pass all the test cases dude !!!!

rajesh_kashyap60 + 1 comment bro, this is a discussion forum. Why are you demotivating a begineers ? Anyway I have submitted that code successfully. Thanks for the advice.

fateme_hajizade + 1 comment I used the same logic, it passes most of the tests but I get timeout error.

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity.

Would really appreciate your feedback like, dislike , comment etc. on my video.

intechworx + 0 comments if(b+1 <= n) numList[b+1] -= k;

Why we are substracting?

kashifali2005 + 0 comments It is failing test case 4 to 13. Not sure why

kashifali2005 + 0 comments It is failing test case 4 to 13. Can you please check

ibirite + 0 comments [deleted]The_Dark_head + 0 comments `long tempMax = 0; long max = 0; for(int i=1; i<=n; i++) { tempMax += numList[i]; if(tempMax > max) max = tempMax; } What is this code doing , why we cant use numList.Sum()`

francisco_rapha1 + 0 comments Used the idea o modifiers, but without the n-sized array. Also in C#. Complexity is O(m log m), that can be less than O(n).

`static long arrayManipulation(int n, int m, int[][] queries) { long max = 0; SortedDictionary<int,long> fakeArray = new SortedDictionary<int,long>(); // 'm' times for (int i=0; i<m; i++) { int a = queries[i][0]; int b = queries[i][1]; int k = queries[i][2]; // bit optimization: a '0' valued query will make no difference on resultant ones if (queries[i][2] <= 0) continue; if (!fakeArray.ContainsKey(a)) { // O( log m ) fakeArray.Add(a, k); // O( log m ) } else { fakeArray[a] += k; // O( log m ) } if (b < n) { b++; if (!fakeArray.ContainsKey(b)) { // O( log m ) fakeArray.Add(b, -1 * k); // O( log m ) } else { fakeArray[b] -= k; // O( log m ) } } } long current = 0; foreach(long modifier in fakeArray.Values){ // O( 2*m ) current += modifier; if (current > max) max = current; } return max; }`

HarishTati + 1 comment *(a+c)+=g; if(d+1<=n){ *(a+d+1)-=g; }

`please explain what's going on here.`

gomesfausto + 0 comments This is a very elegant and beautiful solution.

harshithakamma + 1 comment could you please explain me the logic?

Kanahaiya + 0 comments Hi,

Most of the peope solved this problem but time complexity of your solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

I have created a video tutorial for you and uploaded the same on youtube. Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

Nilesh44 + 1 comment pls explain logic

Kanahaiya + 0 comments Hi,

Most of the peope solved this problem but time complexity of solution is O(n*m) (due to two nested for loops)which can not be used to solve this problem for given time constraint, so you need better approach which beats O(n*m)

I have created a video tutorial for you and uploaded the same on youtube with complete explanation along with code complexity analysis.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

mario89 + 0 comments [deleted]lazy_panda + 1 comment can you please expalin the code.

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

brakesh581 + 0 comments [deleted]theonelucas + 0 comments Will fail if max result is negative

chhayaprakashj + 1 comment i cant pass some test cases, those with very long inputs, i dont know what to do! here is my code: -

long arrayManipulation(long long int n, vector> queries) {

`vector<long long int> v; vector<long long int>v2; long long int max; for(long long int i=0;i<n;i++){ v.push_back(0); } for(long long int i=0;i<queries.size();i++){ for(long long int j=0;j<queries[i].size();j++){ v2.push_back(queries[i][j]); } for(long long int k=(v2[0]-1);k<v2[1];k++){ if(v[k]!=0){ v[k]+=v2[2]; } else{ v[k]=v2[2]; } } v2.pop_back(); v2.pop_back(); v2.pop_back(); } max=*max_element(v.begin(), v.end()); return max;`

}

Kanahaiya + 0 comments hi,

I am sure this comment will definately help you.

https://www.hackerrank.com/challenges/crush/forum/comments/570284

edu_karthikch + 0 comments No need to explicitly assign 0 to every item in the array. malloc() does that automatically.

apocarteres + 1 comment calloc would be faster i guess instead of malloc + zeroing the mem

Mirouhml + 1 comment Could you explain more, please?

apocarteres + 1 comment in the code there are two steps to acquire a new array filled with zeroes:

- At first step the allocation itself: int *a=(int
*)malloc(sizeof(int)*(n+1)); - Second step is to propagate zero values using for-loop upon newly created array.

In my opinion, we can do the same with single call of "calloc". It supposed to be faster in theory, cause calloc is hardware dependent and probably we don't have to loop over N items again just for get them zero valued, since calloc could know is it actually required or not, so on some harware it could be already zeroes or hardware knows better how to zero during allocation faster

Mirouhml + 0 comments okay thanks for the information!

- At first step the allocation itself: int *a=(int

rnj01 + 0 comments This does not work for the current version of this problem. The prompt is asking for the solution of a method with two args, one n for the size ofthe array and a queries array of arrays with start index, end index, and value. There should be no need for gets.chomp().

codesnik + 0 comments And you can easily make solution "sparse" to conserve memory on big arrays, just replace

arr = Array.new(N + 1, 0)

with

arr = Hash.new(0)

and

arr.each do |value|

with

1.upto(N) do |n| value = arr[n]

julie_larson + 1 comment This worked, but I'm still trying to understand why this was the correct answer.

mishyn + 1 comment Because it takes less time to execute. Otherwise solution execution got time outed

victortang_com + 1 comment You are not answering the question.

yanikjay1 + 1 comment I believe what he's trying to say is this: There are two approaches here - 1. The difference array approach (the one every one is praising as being more efficient) 2. The intuitive approach - i.e., just going through each group of a,b,k values and incrementing elements located between a and b in the array, then finally searching through for a max)

The reason approach 1 is more efficient is because the operation for the difference array only requires a maximum 2 adjustments per transformation (you increment the value at the start index by k, and decrement the value after the end index by k). Contrast this with approach 2, where actually going through the array and incrementing every value within the specified 'a-b' range could result in N operations.

So approach 2 could take a max of O(N * M) time- where 'M' is the number of operations, and N is the size of the array And approach 1 could take a max of O(2 * M) time, which is considered equivalent to O(M)

Does that make sense? Someone correct me if I'm wrong! Cheers :)

m_zayed + 1 comment Yeah, if we went with the brute force solution with two nested for loops, we will get a graph as below for the array if we used the data below

arrayManipulation(40,[[19, 28, 419], [4, 23, 680], [5, 6, 907], [19, 33, 582], [5, 9, 880], [10, 13, 438], [21, 39, 294], [13, 18, 678], [12, 26, 528], [15, 30, 261], [8, 9, 48], [21, 23, 131], [20, 21, 7], [13, 40, 65], [13, 23, 901], [15, 15, 914], [14, 35, 704], [20, 39, 522], [10, 18, 379], [16, 27, 8], [25, 40, 536], [5, 9, 190], [17, 20, 809], [8, 20, 453], [22, 37, 298], [19, 37, 112], [2, 5, 186], [21, 29, 184], [23, 30, 625], [2, 8, 960]])

but if we used the second method which is adding at the first index and subtracting at the index afer the last we get this graph

and then we can get the maximum out of summing the results as below

perfectcodder + 2 comments but I can't understand the logic of least step ( summing step )

yanikjay1 + 2 comments the best way to understand it is form a simple example.

say there are 4 of us in a line: 1. me 2. you 3. bob 4. sue and we all start out with zero points.

Thcan be represented as (where my points are in the first index, your in the second, bob's in the third, sue's in fourth):

0 0 0 0

Furthermore, we go through rounds, where in each round a contiguous block of us can receive some set amount of points.

So in round one, say 2 points are awarded to anything in the range of start index = 1, and end index = 2. This means that you and bob, who are in the range, get 2 points.

But rather than write the current score as: 0 2 2 0

We instead want to write the score as:

0 2 0 -2

Because we want each value to represent how much greater/less than it is from the previous-index value. Doing this allows us to only ever need to change two elements in the list for a given round.

Now say we play one more round, and 1 point is awarded to all people in range of index = 0 to index = 1. This gives you

1 2 -1 -2

All I did was add the point value to the start index, and subtract it from the "end index + 1".

Then calculating the max is simply a matter of going through that list, adding each element to an accumulator variable (as this summing process reveals the actual value of an element at any given point - e.g., you would have 3 points at the end because your score is the result of 1 + 2), and having a variable which keeps track of the running max.

jie_lyu + 0 comments Thank you for your detailed explanation. It helped!

JDODD74 + 0 comments Because we want each value to represent how much greater/less than it is from the previous-index value.

This is the best explanation of the difference array I have read in these discussions -- thank you for making it click in my head.

Kanahaiya + 0 comments Hi,

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

thatsamorais + 1 comment I would not suggest eclipsing

`list`

ShubhamV06 + 2 comments Can some one please help as to why my solution gives a segmentation fault?

# include

# include

# include

# include

int main() {

`unsigned long long int n,m,l,b,k,i,val=0; scanf("%llu%llu",&n,&m); unsigned long long int a[n+1]; for(i=1;i<=n;i++) { a[i]=0; } while(m--) { scanf("%llu%llu%llu",&l,&b,&k); for(i=l;i<=b;i++) { a[i]+=k; if(a[i]>val) { val=a[i]; } } } printf("%llu",val); return 0;`

}

dr_cooper + 1 comment Maximum array size has to be 10^7 which is not possible in case of C. I tried Dynamic memory allocation (malloc) which worked but got TLE for bigger test cases

ShubhamV06 + 0 comments yeah it is getting a tle. We need to use a different algorithm. I wanted to know what the problem in my code was so i posted my solution.

vicky_20 + 0 comments use long instead of int.

PratikDreamer + 1 comment Hi Vineet could u explain what logic have you followed?

Kanahaiya + 0 comments Hi,

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

timkay + 5 comments Same solution but optimized:

from itertools import accumulate n, m = map(int, input().split(' ')) dx = [0] * (n + 1) # allow run past end for _ in range(m): a, b, c = map(int, input().split(' ')) dx[a - 1] += c dx[b] -= c print(max(accumulate(dx)))

kai_gowers + 0 comments That's so good! You are awesome!

ygrinevich + 1 comment can you explain why it's (n+1) and not (n)?

Jellofish + 0 comments The question asks for a 1 indexed list and the a,b values read in are read in starting from 1 not 0. If you do not use (n+1)if b happens to be the last number of your list it will go out of bounds.

The 0 index value will always be 0 so it doesn't affect your answer when you sum for max difference.

kunalwww1994 + 1 comment why are you subtracting dx[b] -= c ?????

argsv + 0 comments Walk through the array it may help.

Query 1 -> [1, 2, 100] Array 1 -> [0, 100, 0, -100, 0, 0, 0] Query 2 -> [2, 5, 100] Array 2 -> [0, 100, 100, -100, 0, 0, -100] Query 3 -> [3, 4, 100] Array 3 -> [0, 100, 100, 0, 0, -100, -100] Array Accumulated -> [0, 100, 200, 200, 200, 100, 0] Result -> 200

vietanhtran_96 + 0 comments The use of accumulate is so brilliant!

mofo_jones + 0 comments Nice

tug16183 + 0 comments How does that bottom for list return the maximum value in the list? Could you explain the logic to me please?

sonic88bt + 0 comments Why are you checking if "y" is <= len(list) ? This is given by the problem, right?

AamirAnwar + 0 comments Really liked this solution! One question, shouldn't it be max=-float('inf')?

P.S - Talking about the python version

gg_germain + 0 comments Just improved your code a bit. - split() is slightly faster than split("") - an if was not needed as always true - variables names fit those used in the exercise

if __name__ == "__main__": n, m = [int(n) for n in input().split()] l = [0]*(n+1) for _ in range(m): a, b, k = [int(n) for n in input().split()] l[a-1] += k l[b] -= k; max = a = 0 for i in l: a+=i; if max<a: max=a; print(max)

Etiennera + 0 comments [deleted]GhostlyMowgli + 0 comments I took the ideas explained here to put also into python in a bit more spelled-out way to help make sense of this more abstract work-around.

`def arrayManipulation(n, queries): # Big O (N) res = [0]*(n+1) # we only really need one terminal row, since we're applying each pass to all rows below # loop through all the queries and apply the increments/decrements for each # Big O (M) (size of queires) for row in range(len(queries)): a = queries[row][0] b = queries[row][1] k = queries[row][2] res[a-1] += k # increment the starting position # this is where a loop would increment everything else between a & b by k # but instead of taking b-a steps, we take a constant 2 steps, saving huge on time res[b] -= k # decrement the position AFTER the ending position # now loop through res one time - Big O (N) (size of res) sm = 0 # running sum mx = 0 # maximum value found so far for i in range(len(res)): sm += res[i] if sm > mx: mx = sm # total run time is Big O (2*N + M) >> Big O(N) return mx`

The key concepts in my opinion here are:

1) we don't need to build the full aray, since it's impossible for any row but the last to have the max value. This is impossible because we apply the adjustments to every subsequent row of the resulting 2-D array.

2) we don't need to actually increment each value for indices 'a' through 'b'. While that's the most straight-forward way, that also requires x many (a minus b) steps for each pass of the loop. By noting instead where to start incrementing and where to stop incrementing (noted by decrementing after what would be the final increment), we can note the adjustments to apply without having to take every step each time. We can then run a separate single loop to go over each of the increments and keep a running sum of all the overlapping increments. The decrement allows us to know when that range where the increment applied is over by reducing the running sum by that number - in other words when that range is ended, we would have to look at overlapping increments excluding that one that just ended going forward to see if anything is bigger.

Someone else in here gave an image of a stair/hill which I found extremely helpful in visualizing and understanding this concept. The basic idea here is that instead of actually applying each increment, we can just note what the increment and range is and one by one go through each place and apply all the compounded increments at once. Loop 1 saves the increments in a different format, Loop 2 checks the overlap. And by using two separate loops we have basically Big O (N) rather than Big O (N^2) - or more specifically Big O (2N + M) instead of Big O (NM + N), where N is the size of the resulting array and M is the size of the queries array.

karankhajuria1 + 0 comments def arrayManipulation(n, queries): arr=[0]*n for _ in queries: start=

*[0]-1 end=*[1]-1 val=_[2] topVal=end+1 for i in range(start,topVal): arr[i]+=val return max(arr)m = input().split()

n = int(nm[0])

m = int(nm[1])

queries = []

for _ in range(m): queries.append(list(map(int, input().rstrip().split())))

result=arrayManipulation(n, queries)

mercy1 + 0 comments not sure why, but when I run your code as it is I get an error: if((y)<=len(list)): list[y] -= incr; IndexError: list index out of range

but when I do run that line like this:

if((y)<=len(list)): list[y-1] -= incr;

that works

radialapps + 0 comments A bit less efficient, but more readable

def arrayManipulation(n, queries): a = [0] * (n + 1) for q in queries: a[q[0] - 1] += q[2] a[q[1]] -= q[2] for i in range(1, len(a)): a[i] += a[i-1] return max(a)

m_zayed + 0 comments [deleted]m_zayed + 3 comments Python 3 more clear code

def arrayManipulation(n, queries): my_array = [0] * (n+1) count = 0 temp = 0 for first,last,value in queries: my_array[first-1] += value my_array[last] -= value for item in my_array: count += item if count > temp: temp = count return temp

perfectcodder + 2 comments can you explain your logic ??

brayoni_ian + 0 comments [deleted]Kanahaiya + 0 comments Hi,

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

brayoni_ian + 0 comments Thank you. This is clear.

You could also find the max using:

`max(accumulate(my_array))`

See below:

def arrayManipulation(n, queries): # initialise list arr = [0] * (n + 1) # update for a, b, k in queries: arr[a - 1] += k arr[b] -= k # accumulate values return max(accumulate(arr))

However, I have not understood the theory behind the

*difference array algorithm*, why it works the way it works and how to derive it.abhipatel1 + 1 comment can anyone debug this program

bhavyseth + 0 comments [deleted]

rounak077 + 1 comment Brilliant man, mine was working for initial test cases but getting timeout for rest.

def arrayManipulation(n, queries): a = [0]*(n +1) z=[] for i in queries: z.append(i) for i in range(len(z)): for j in range(int(z[i][0]),int( z[i][1])+1): a[j] = a[j]+int ( z[i][2]) return max(a)

Kanahaiya + 0 comments Hi,

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

h1017691668 + 0 comments Thanks , max(list) wil be simpler

navya_lucky03 + 1 comment can the same be written like this: def arrayManipulation(n, queries): arr_n=[0]*n for i in range(len(queries)): n1=queries[i][0] n2=queries[i][1] for j in range(n1-1,n2): arr_n[j]=arr_n[j]+queries[i][2] return max(arr_n) This is giving me a tiemout error while submitting. can u assist here?

h1017691668 + 0 comments yeah,at the beginning, l got the same problem as well. This algorithm's complexity could be o(n). Try to learn something about prefix sum array.I hope this can help.

blauge_reskrooge + 0 comments But when y=len(list), list(y) will throw an error right? out of index error?

alvaroSanchez + 1 comment Another python translation:

def arrayManipulation(n, queries): arr = [0]*(n+2) for a, b, k in queries: arr[a]+=k arr[b+1]-=k result = acc = 0 for x in arr: acc+=x result = max(result, acc) return result

dimitrantz + 0 comments Brilliant!

harishmallepall1 + 1 comment can u explain why we are subracting i dont understand the if statement ......

Kanahaiya + 0 comments Hi,

I have uploaded a video tutorial which can help you to understand the problem as well as logic.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

georgedubuque + 0 comments This breaks when

`y == n`

. The if when decrementing`list[y]`

should be`if y < len(list): list[y] -= incr`

. Overall great solution! Thanks for doing it in python!madhumaa2000 + 0 comments Excellent, but why you are using this step.. list[y] -= incr; and that "if" condition is always correct, right..! Can you explain please...

steven_succar + 0 comments [deleted]abhimanyuaj001 + 0 comments for some test cases its showing teminated due to timeout what can i do

def arrayManipulation(n, queries): a=[] for i in range(n): a.append(0) for j in queries: for k in range(j[0],j[1]+1): a[k-1]=a[k-1]+j[2] return (max(a))

shireen305 + 1 comment these solutions give runtime error

abhimanyuaj001 + 1 comment yup for 7 testcase its showing time out what should I do??

Kanahaiya + 0 comments hi,

I am sure this comment will definately help you to solve your problem.

https://www.hackerrank.com/challenges/crush/forum/comments/570284

nivdatta88 + 3 comments I still have some doubt if anyone could explain it. I understand that it is making use of difference array to keep track of the difference between items rather than updating every item.

However as I understand, size of difference array is one less than total items in the actual array.

So as per the demo program given,

ARR DIFF (5,3) 0 0 0 0 0 0 0 0 0 (1,2) 100 100 0 0 0 0 -100 0 0 (2,5) 100 200 100 100 100 100 -100 0 0 (3,4) 100 200 200 200 100 100 0 0 -100

However the logic used seems to be a variant of difference array that I am not able to understand. It would be great if someone could help me connect the dots from this difference array to the actual working solution of this problem. Thanks.

zeralight + 1 comment Hello, I don't really get your DIFF representation, but here is the deal:

you need to keep in mind that at the end of queries to get the real value of i we need to sum the values of A[0]->A[i].

for a query (l, r, x) if we add x to A[l] then we're sure later for every k in [l, r] sums(0->k) will include x.

the problem is that every k > r will also include that x, therefore we'll decrement l[r+1] by x.

Now the sum through 0 -> k will include x only if k >= l and will exclude it only if k > r which is equivalent to:

sum(0->k) will consider x only if k is in [l, r]

d_niladri95 + 0 comments [deleted]

julie_larson + 3 comments I think the DIFF matrix should be:

`0 0 0 0 0 100 0 0 -100 0 0 100 0 0 0 0 0 100 0 -100`

then x is 100+100+100-100-100 = 200

still don't quite get it though

cse160001033 + 0 comments [deleted]cyxingfa + 1 comment I think the Diff should like this

0 0 0 0 0 100 0 -100 0 0 <--- 1 2 100, 3rd col is 100 less than 2nd col 100 100 -100 0 0 <--- 2 5 100, 2nd col is 100 more than 1st col 100 100 0 0 -100 <--- 3 4 100, now 3rd is equal to 2nd, and 5th is 100 less than 4th col

so finally, 1st col is always the real value, and others are accumlated of previous values and it self. so actual values are 100 200 200 200 100, max is 200

marc_ponchon + 0 comments Thanks this helps me to understand how it works.

feynman65 + 2 comments @julie_larson

100+100+100-100-100 = 100 still not 200 I've been banging my head still not able to understand this solution.filipetrm + 0 comments The final diff array should be like this:

100 100 0 0 -100 (straight foward from the code)

So when you iterate over it to get the accumalated sum you'll get this:

iteration value of x accumulated sum 0 0 0 1 100 100 2 100 200 3 0 200 4 0 200 5 -100 100

The maximum value of the accumulated sum is 200.

joshumcode + 0 comments They're always checking the sum against the count, so when the sum is less than the count (which is the greatest sum so far) the count doesn't update.

nayeemnawaz93 + 0 comments As i have understood (5 , 3) 0 0 0 0 0 (1 , 2) 100 100 0 0 0 -> first and second places i.e a=1,b=2 , k=100, now a+k = 0+100 = 100, b+K = 0+100 = 100. (2 , 5) now from 2 to 5 i.e 2,3,4,5 are added with K i.e 100. already 2=100 now 2 becomes 100+100 =200, simialrly 3 = 0+100=100,4=0+100=100,5=0+100=100. (3 , 4) now 3 and 4 are added with 100. ie 3 value is 100 it becomes 100+100=200, 4 becomes 100+100=200.

among these 200 is max value.

cse160001033 + 4 comments instead of writing long int *a=long int[n+1] if I write long int a[n+1]; It shows segmentation fault after test case 7. why? Explain please.

Alex_leo + 0 comments yes ,the same with you, i am confused

julie_larson + 0 comments you aren't allocating space for

**a**that way. He uses**new**to create an array on the heap. You are trying to declare it on the stack, but you can't do that with a dynamic value like 'n'ami3443 + 2 comments As per my knowledge, u can create an array to size upto 10^6 in stack and for size above it we need to create in heap so we do it dynamically. https://discuss.codechef.com/questions/28604/maximum-size-of-an-array refer this

ShubhamV06 + 0 comments Thanks @ami3443

kousiknandi + 0 comments Nice explanation.

RuraDund + 0 comments If you have a problem understanding dynamic memory then make use of vector . It allocates memory dynamically wherein the user doesn't have to deal with pointers .

vector arr(n); declares an array of long int of length n.

nivdatta88 + 0 comments [deleted]sandeep007734 + 0 comments Thanks for explaining the logic behind this.

tanyaangra123 + 1 comment Here ,I add all the positive values of the difference array to get the maximum element . But , the answer comes out to be wrong for all the test cases except sample one ...

prakash_katudia + 0 comments Use long int in place of int.

manishch1095 + 1 comment what do u mean by previous element? is it a[p-1]for a[p] or the previous state of a[p] before the current updation operation? difference is betweentwo succesive elements or in the same element before and after one query is executed?

tanyaangra123 + 0 comments It is a[p-1] for a[p] ie. the difference between two successive elements.

Well ,now ,i understood my mistake . I was just adding the positive differences. But the problem requires maximum element in an array , so there must be a comparison while adding the differences.

Well thanks!

tanujindal1989 + 0 comments wonderful explaination...Thanks a lot

chaizhongzhong + 0 comments Hi gabriel

chikmid + 0 comments how is element at b+1 the previous element?

quietwalter + 0 comments Thank you ! Good to see this better way to store the difference only , so array size = n+1 only !

Kanahaiya + 0 comments Hi,

I have created a video tutorial for all and uploaded the same on youtube with complete explanation along with code complexity analysis.

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

michellerodri247 + 0 comments [deleted]alexandis + 0 comments [deleted]

shivamckaushik + 1 comment Your Explanation is wrong and misleading.Logic behind @ amansbhandari algorithm is at any index p he is storing by how much this value is more than previous value, subsequently if value from p to q is 100(to be exact 100 more than other values) q+1 should be 100 less than values between p and q.That is why we subtract 100 at q+1 index of array.

adi928 + 0 comments That's it. Thanks for a simpler explanation.

akashjus + 2 comments The solution works, but how did you came to this theory ?

guangzefrio + 6 comments That's what I understand this theory. We can try to understand this logic like we imagine Supermario walking on a N width horiz line. a and b is the point on the line, k is the mushroom Mario like to eat. When Mario go to the point at a,he eat the k size mushroom and become taller,after he have walked through point b, his height reverse to the origin height before he eat the mushroom. eg. 1. When Mario is walking to a, he eat a k size mushroom, and become k bigger 2. Then Mario is walking to a', he eat a k' size mush, and become k' bigger, now Mario's height is (k + k') 3. If Mario have walked to b, so he pooped out the mushroom and become k smaller, the only way that he can become larger is to meet a new (a,b) point and eat a new k size mushroom 4. The rest can be done in the same manner.

`What we need to do is tracing the Mario's biggest height when walking through that muliple query's a and b point.`

ayu111 + 1 comment awesome explanation.. btw i wanted to ask if this is what we call segment tree..and if we can use this method to solve questions with the segment tree tags.. P.S. : I have a very little knowledge about the segment trees.

Thanks in advance.

nikhiljain1222 + 1 comment you can refer top coder or either gog to understand the concept of segment tree. ;)

ayu111 + 1 comment acha bhai :D

nikhiljain1222 + 0 comments haha.. ;)

vventura + 0 comments This is literally the only reason I actually was able to understand the logic of the code. Thank you for that, seriously.

leong1568 + 0 comments Very nice explanation. The explanation in the question does misleading on how to implement the solution.

rawshar_bhardwaj + 0 comments Thanks a lot for your explanation, it really simplified the problem. :)

rhuang23 + 0 comments ahaha great explanation, thanks so much!

qwer1234spam + 0 comments The only explanation I understood

Kanahaiya + 0 comments Hi,

Here is the video tutorial for my solution O(n+m) complexity passed all test cases.

Would really appreciate your feedback like, dislike , comment etc. on my video.

vaib95 + 0 comments i dont get why did you subtract from q+1th element?

gauravkochar_gk + 0 comments sorry dear ,I dont get it .Plz tell me, how do you think about this logicc....

sarbajitnandey + 0 comments Check your code with this inputs

*i/p:*7 2 1 5 100 2 3 500[deleted] + 0 comments here we are adding value only in a[p] but we have to add value from a[p] to a[q] .and what is the requirement of a[q+1] here????? plz clear my doubt ...

rghvmandowara + 0 comments how you thought this ?????

piyushkushwah131 + 0 comments why are we substracting from q+1? anyway how one can think to this level?

Anirudh_K + 1 comment Why is my code not working on the same approach ? Please help Can't find anything wrong !

# include

# include

using namespace std;

int main(){

`int N,M,a,b,c; cin>>N>>M; long int *arr=new long int[N+1]; for(int i=0;i<M;i++){ cin>>a>>b>>c; arr[a]+=c; if(b+1<=N){ arr[b+1]-=c; } } int x=0,ans=0; for(int i=1;i<=N;i++){ x+=arr[i]; ans=max(ans,x); } cout<<ans;`

return 0; }

mtkocak + 1 comment keep an eye on edges.

Anirudh_K + 1 comment Still No Clue!

vadim_tarasov + 1 comment Try to change

for(int i=1;i<=N;i++){

to

for(int i=0;i<=N;i++){

guangzefrio + 4 comments We can try to understand this logic like we imagine Supermario walking on a N width horiz line. a and b is the point on the line, k is the mushroom Mario always like to eat. When Mario go to the point at a,he eat the k size mushroom and become taller,after he have walked through point b, his height reverse to the origin height before he eat the mushroom. eg. 1. When Mario is walking to a, he eat a k size mushroom, and become k bigger 2. Then Mario is walking to a', he eat a k' size mush, and become k' bigger, now Mario's height is (k + k') 3. If Mario have walked to b, so he pooped out the mushroom and become k smaller, the only way that he can become larger is to meet a new (a,b) point and eat a new k size mushroom 4. The rest can be done in the same manner.

`What we need to do is tracing the Mario's biggest height when walking through that muliple query's a and b point.`

Ea9Le + 0 comments Best explanation, thanks.

adarshawasthi851 + 0 comments thanks a lot bro this is best explanation i found of this problem.

dania_dlbani + 1 comment This is the best explanation ever. Thank you.

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