apghr A compact C++ solution:
#include <iostream> int main(){ int c, n, max = 0; std::cin.ignore(); while(std::cin >> n) max < n ? c = !!(max = n) : c += max == n; std::cout << c; return 0; }
v1nea Like this a lot. Thanks.
mohammedavez85 `c
include
include
include
include
include
include
include
int birthdayCakeCandles(int n, int ar_size, int* ar) { // Complete this function int i,j,temp,count=0,b; for(i=0;i
b=ar[0]; for(i=0;i<n;i++){ if(b == ar[i]){ count++; } }return count; }
int main() { int n,ar_i; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n, n, ar); printf("%d\n", result); return 0; } BRO THIS CODE SAYS TIME LIMIT EXCEEDED PLEASE HELP
Yakubk_98 use c++ and you'll be fine
int birthdayCakeCandles(int n, vector <int> ar) { // Complete this function int max = ar[0]; int count = 0; for(int i=0; i<n; i++) if(ar[i] > max) max = ar[i]; for(int i = 0; i < n; i++) if (ar[i] == max) count++; return count; }
sandeepkataria this code was working on some test case but not all please help me
vaishaligarg2012 /*You should do like this... */
int max=ar[0]; int count=0; for(int i=0;i<n;i++){ if(ar[i]>max){ max=ar[i]; } } for(int i=0;i<n;i++){ if(ar[i]==max){ count++; } } return count;
jdanielys07 In java, maybe this way can help you! :)
if(ar!=null && ar.length>0 && n>0) { int maxHeight=0, numMaxHeight=1; for(int i=0; i<ar.length; i++) { if(maxHeight<ar[i]) { maxHeight=ar[i]; numMaxHeight=1; }else if(maxHeight==ar[i]) { numMaxHeight++; } } return numMaxHeight; } return 0;
Stelios10 just an optimization, even though in this case your code is fine: if you try an array with negatives, then you ll never get inside your if-statements. so the numMaxHeight will never change, so you dont really compute it in such cases. so this solution for an array with negatives, where the max negative appears more than once, will fail. eg [-1, -1]
proposed solution: set numMaxHeight = 0 on initialization
shashank_sm [deleted]sirmaridvan Can a birthday cake candle have a negative length?
Zachurnia_Tate No, candles cannot have a negative length buddy. I don't kniw what kind of world your living buddy.
salvgal Buddy, even though your real world experience gives you a meaning to things, you have to cover up your code from all the edge cases: you never know what kind of input your function take, especially if the problem description does not specify that the input array contains positive integers only, and a negative ints array is a valid input for our function
atineo You are correct that you have to proof safe your code but the mandate is telling you what input you are getting. Positive integer arrays. Refer to the constraints section of the explanation.
szatmary_zoltan1 But you shouldn't do anything like int max = 0 (except if there are no negative numbers or if it's indicating the index of the maximum value). Instead you should use: int max = ar[0];
shrikanth_n23 You can simplify it like this:
int max=ar[0]; int count=0; for(int i=0;imax){ max=ar[i]; count = 1; //Reinitialize count if you found a new max }
else if(ar[i]==max){ count++; }} return count;
I this case you are traversing the vector only once.
pkarif4249816 [deleted]pkarif4249816 [deleted]soumabratabhatt1 i did the same thing.But it is showing me 4 test cases failed.
osama_abusitta int max=ar[0]; this wrang staement , replace with int max= 0
RizviTammar exactly same i did but it give me error of indexoutofbound exception an can't resolve it" :( what to do?
sandhyarani_dot1 [deleted]24god10 this code and your logic so complex, try to optimize it, you can sort this vector and count the element(s) had duplicate with the last( if you sort it greater).
pa980403 I did like this as well but still it is not passing many test cases are not
Its_NM_ Its the same thing
sensmaduranga its not working
jord_fumar Be aware of time complexity. In my testcase 4, n was 10000. If you use a sorting algorithm with n^2, you will get a runtime error.
The way I structured my code is:
- Sort the array from min to max (quickSort)
- Parse from the very end of the array and iterate count++.
- When your values are no longer identical, stop.
jagmeetsingh I tried this
static int birthdayCakeCandles(int[] a) { Arrays.sort(a); //double pivot QuicKSort int count = 0; int i = a.length - 1; int tallest = a[i]; while (a[i--] == tallest) { count++; } return count; }
It is failing the test case
100000 followed by 100000 times 9999999 (as height of candles(array values)).
x_ireesh_x Pass all Testcases..
static int birthdayCakeCandles(int[] ar) { Arrays.sort(ar); int result = 1; int decriment = 2; while((decriment <= ar.length) && (ar[ar.length-1] == ar[ar.length-decriment])){ result++; decriment++; } return result; }rishavtandon93 Try This
static int birthdayCakeCandles(int[] ar) {
int count =0; Arrays.sort(ar); int maxHeight = ar[(ar.length)-1]; for(int i=0;i<ar.length;i++) { if(ar[i]==maxHeight) { count++; } } return count; }
Kanahaiya Hi,
Your solution is good but it will take O(nlogn) time due to sorting which can be optimized to O(n)
Here is the video explanation of my solution in O(n) time -
Any comments and feedback will be appreciated.
munnidivya93 thanks
Amit_Sharma need to add one more check in while condition for all the array elements with same value
while ( i > -1 && a[i--] == tallest ) { count++; }
dpkcareergreatn1 Count should be initialized to 1 as there is atleast one tallest. In this case, if the frequency of largest element is 0, count would return 0 which is an error
harikrishnaprak1 while (i>-1&&a[i--] == tallest) { count++; }
You just forgot to check whether the index i is within the bounds.
bertgayus if you use quicksort u already have O(nlogn) which is worse than just iteratring through the array and saving the highest value and then just iterate a second time incrementing a counter everytime the value is similar to the saaved one. then complexity is O(2n) = O(n). With n = 100 you would have ~ 600 steps(n*logn = 100 * 6 = 600) with your solution and just 200 steps with mine
matthew_zilkie Did you try using c++ std::sort(ar.begin(), ar.end()); ? I did and it worked flawlessly. int birthdayCakeCandles(int n, vector<int> ar) { // Complete this function sort(ar.begin(), ar.end()); int max = ar[ar.size() -1]; int count = 0; for(auto num : ar) { if(num == max) ++count; } return count;}
deepuvattikonda Thanks a lot!!.I was just wondering where did i go wrong..Thanks a lot again
luckysam09 ar[i]-max==0 in second if worked for me
generalom1234 sandeep did u use unsigned long or any other data type which allows you to take in large amounts of data?
akshaythear This code is assuming that first entered element is always the largest one.So it wont work in some cases... in short it is a wrong code.
r_umans Since height of the candle is always at least 1, you can safely assume max = 0; Then loop all at once, you dont need a double loop. (Java syntax)
If(max < ar[i] ) { max = ar[i]; count =1; } else if (max==ar[i]) count++;
x_ireesh_x [deleted]x_ireesh_x [deleted]x_ireesh_x [deleted]
Anoop_SS You don't need the second loop. Checking of equal can be done in the else of
if(ar[i] > max)matthew_zilkie [deleted]
mercia_a thank you!
shahzadhafeez63 instead of using two loops, we can solve it in one loop. static int birthdayCakeCandles(int[] ar) {
int size = ar.length, max = Integer.MIN_VALUE ,count = 0; //find tallest candle and count for (int i=0;i<size; i++) { if (ar[i]>max) { max = ar[i]; count = 1; } else if (ar[i]==max) { count++; } } return count; }harikrishnaprak1 Nice!
jacobkalas I came up with the same solution as you but in Java instead, working on all test cases!
josephawilson86 Ask yourself a few questions:
- What do I need to return?
-The number of highest candles
- How would I get that number?
-I need to know what the highest candle is.
-I need to know how many of them there are.
- How would I know which is the highest candle?
-Simple comparison should work for knowing what the highest is. Just store it in a var and reset when I find one that's bigger.
- How would I track how many of they there are?
-Store the quantity in a var, increment it whenever I find one of the Highest candles and reset it each time I find a new highest candle.
salvgal Thank you Joseph, your explanation of the problem brouhgt me to this simple solution that run in O(n):
int ret = 0; int max = 0; for(int a : ar){ if(a > max){ ret = 1; max=a; } else if(a == max) ret++; } return ret;Sometimes the hardest part of these problems is finding what the question is asking more than how to solve it.
wdodman ar.sort(); ar.reverse(); let m = 0; let k = ar[0]; while (parseInt(ar[m]) === parseInt(k)){ m++; } return m; }
passes 8 of 9.
rethangotit4 return ar.count(max(ar))
qu4ku i'm not into c++, but this looks savage :)
trungskigoldberg Can you explain more about the complicated, savage version of that ternery statement? I really want to know this nifty code!
apghr while(std::cin >> n) max < n ? c = !!(max = n) : c += max == n;
can be translated as:
while(std::cin >> n){ // for each candle n check: if(max < n) { // does n set a new record in height? max = n; // if that's so, then n is the new max height c = 1; // and the counter c must be set to 1 again. "!!" is a cheap trick to convert any value different from 0 into 1 } else { // otherwise check if the new candle is as tall as max if (max == n) c++; // in that case, add 1 to counter (otherwise add 0) } }
arpita414 Too good
oswald200801 Nice one !!
aishwarya217 Good!
adityabcs93 nice one
lehieuminh231 nice. thank you
tarsum Well I came up with similar algorithm, so thumbs up. Also seems like solution in editorial goes though array twice, and this goes only once.
Vishwas92 hats off..
ichidan Nice one. I arrived at same algo in C, minus the one liner ternary. Looks neat, I like it. Although I'd be curious if there is not a small performance loss of doing:
c = !!(expr);
vs
(expr); c = 1;
the former may require additional conditional logic depending on architecture (for instance x86 - http://riffwiki.com/MOV_(x86_instruction) - the MOV instruction doesn't affect ZeroFlag, so I'd guess CPU would have to evaluate whether your expr results in a non-zero value. It's obvious to us as we can see the wider context of the algorithm, but not the CPU)
kurian_benoy thanks
viigihabe It is not CPU that analyses your c++, the compiler does. CPU doesn't "thin" but compiler does. It is really pointless to "optimize" your code this way because compiler is required to emit assembly that does what is ment but in which order and through which instructions, it is not specified. You better belive it does good job. You express c++ although it is slower than ++c: there is no way the compiler would use slower version if it doesn't afect the program flow. With no optimization flags may be.
lakshay91dutt [deleted]
rahulsahu_blog This is brilliant!
gharibakhamis nice why didn't i think of this sooner
parvpareek TYSM!! Helped me a lot..
sabahat_usman_su I did the same thing but my code fails for some tests :O
jagmeetsingh I tried same in java. It worked! Thanks (:
int temp, max = 0, result = 0, n; for (int i = 0; i < arCount; i++) { n = scanner.nextInt(); temp = (max < n) ? (result = ((max = n) >= 0) ? 1 : 1) : (result += (n == max) ? 1 : 0); }
Ray1984 Very nice -- thanks for nudging me to think about it as a streaming algorithm :-)
colonel_bishop It's cool! Thanks!
ScienceD I dont understand how cin.ignore(); works... And how you can feed input into loop condition...
thermosphere453 cin.ignore() ignores the input of number of candles as its not needed in his algorithm and the input in loop works as long as the input isnt 0 or null, while loop works for any number in the test condition.
kachi_kinjal can u give the solution for c programming.
oswald200801 int main(){ int n; scanf("%d",&n); int *height = malloc(sizeof(int) * n); int max=height[0]; int count=0; for(int height_i = 0; height_i < n; height_i++){ scanf("%d",&height[height_i]); } for(int i=1; imax){ max=height[i]; } } for(int k=0; k
unicpawan can u please tell me why this code is not able to pass all the test case
// the code goes here..
int main(){ int n,i,max=0,frequency=0;
int ar =(int) malloc(sizeof(int) * n);
scanf("%d", &n); for( i = 0; i < n; i++){ scanf("%d",&ar[i]); } for(i=0;i<n;i++){ if(ar[i]>max){ max=ar[i]; frequency=1; } else if(ar[i]==max){ frequency++; } } printf("%d",frequency); return 0;}
bharat_reddy18 include
include
include
include
include
include
include
int birthdayCakeCandles(int n,int a[]) { long long int i,max=0,d=0; for(i=0;imax) { d=1; max=a[i]; } else if(a[i]==max) { d++; } } return d;
}
int main() { int n; scanf("%i", &n); int *ar = malloc(sizeof(int) * n); for(int ar_i = 0; ar_i < n; ar_i++){ scanf("%i",&ar[ar_i]); } int result = birthdayCakeCandles(n,ar); printf("%d\n", result); return 0; }
kuppesh_ds Return frequency to calling function instead of printig here.
vishalgwalani6 if you have case like 20,40,90,90,90 so you see according to your code in max variable value will 20 and frequency will be 1 and after the loop ends your result will be 4 instead it should be 3
shac_bandagi your else if part should be in for loop and in if statement remove that frequency set otherwise max number counted twice.
anurag0430 your are returning count of candles of same height. We have to return the number of candle he/she can blow out.
gaurav_877 you havent calculated the final max value and processed it before hand for comparision
maheshmnj what is max before entering the for loop??
gjaiswal108 because you have allocated memory to pointer ar using malloc function before taking the value of n from user. you should first take input n from user and then allocate memory to the pointer, i.e. first use scanf("%d",&n); then write int ar=(int)malloc(sizeof(int)*n);
hope, you understand.
kmschaal902 compact indeed, great job!
andrew_hedy77 Yeah, yours is way cleaner than mine
int birthdayCakeCandles(int n, vector <int> ar) { int count = 1; sort(ar.begin(), ar.end()); reverse(ar.begin(), ar.end()); for (int i = 0; i < n; i++){ if (ar[i] == ar[i+1]){ count++; } else break; } return count; }
amitsoni90 Why we need vector here?
cse_115 good solution
wem18 Solid solution. Wouldn't it be more efficient to do a for loop such as
//call sort //set int max = ar[n] for(int i = n; i>0; i++) { if(ar[i]==max) { count++; } else{ break; } return count; }
So you can avoid calling both sort and reverse? Not trying to correct you, genuine question
abhi69sindh it will not pass all the testcases
3001pratap [deleted]
shantanu_knight Great one
karthikeyan90422 What is the use of cin.ignore() please tell me
lyj1186669164 That's so nice.Thanks
VIDYARANIGIDDE what is the role of cin.ignore() ?? what if we do not write that line?
cu_16bcs2416 wao! stupendous code!
benaffleks Can someone explain to me :
max < n ? c = !!(max = n) : c += max == n;
nvinson234 C++ allows for integers to be treated as logical values. If the integer is non-zero, it's "true" and if the integer is zero it's "false". The
!is the logical-not operator and the language standard says!0will be equal to 1.A similar rule holds for
max == n. Whenever, that expression evaluates to "true" it will also represent the numerical value 1.Assignments, such as
max = n, also have the side-effect of "returning" the assigned value (n in this case).Therefore, this is a really terse way of writing:
if (max < n) { max = n; c = 1; } else if (max == n) { c += 1; }
breakbadsp bro use python3 :
def birthdayCakeCandles(ar): return ar.count(max(ar))
No looping, no veriables, no mess
nsuniln1 [deleted]overwhelm wow this is crazy, thank you
lampladam1 i did the same. python is so useful when it comes to counting the number of occurences
dondorian No looping?
So, how were "max" and "count" calculated?
aida71 This is wrong if you do an interview. They most likely want to see a solution which is O(n) and yours is O(n^2).
This is mine, which is O(n):
def birthdayCakeCandles(ar): max_h = ar[0] max_n = 1 for c in ar[1:]: if c > max_h: max_h = c max_n = 1 elif c == max_h: max_n += 1 print(max_n)
georgechristied1 cheers for pointing this out, but is the solution provided by breakbadsp actually O(2^n)?
k_mouratidis No, it's not. Also, if this were an interview I would get extra points. It is actually 2*n (compared to n), so both are O(n). But you must keep in mind that this is Python, not C++. It IS actually faster to use its functions rather than looping through everything. In fact, it is almost 2x faster. Code for benchmarks:
from timeit import timeit from random import randint from funtools import partial f = [randint(0,5) for _ in range(10000)] def breakbadsp(ar): return ar.count(max(ar)) def aida71(ar): max_h = ar[0] max_n = 1 for c in ar[1:]: if c > max_h: max_h = c max_n = 1 elif c == max_h: max_n += 1 return max_n # 1.72946... seconds timeit(partial(breakbadsp, f), number=10000) # 3.10752... seconds timeit(partial(aida71, f), number=10000)
Also, this is Python. Even if the second was faster, you should pick the first for 90% of the cases. You need less time to figure what it's doing, and is less error-prone.
samarththeengin1 U are great bro.
vtrivenirao so simple !!!!
gleventis94 [deleted]marcomereu2015 Thank you!
ilya_zarembsky Cute
Kanahaiya 100 % working code with video explanation -- All Test Case Passed..!!
Here is the video explanation of my solution -
and you can find most of the hackerrank solutions with video explaination here-
https://github.com/Java-aid/Hackerrank-Solutions
and many more needs to be addeed.
Regards,
Kanahaiya Gupta
Git Hub URL | https://github.com/Java-aid/
LIKE US | https://www.facebook.com/javaaid/
SUBSCRIBE US | https://www.youtube.com/c/JavaAidTutorials
TELEGRAM LINK| http://t.me/javaaid
gadarsh555 can you explain it's process of working ?
Kanahaiya You can go through this tutorial-
Here is the video explanation of my solution -
shubhambajare thanks a lot !! This helps me to fingure out the problem. :)
Kanahaiya most welcome. :)
If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.
Do not forget to subscribe my youtube channel for hackerrank solutions .:)
ali75 one line code C++:
return count(ar.begin(), ar.end(), *std::max_element(ar.begin(), ar.end()));
umangsomtiya8083 BRO THIS CODE SAYS TIME LIMIT EXCEEDED PLEASE HELp
Kanahaiya Hi,
your solution may be correct but may not be efficient .you can optimized your solution further.
Here is the video explanation of my solution in O(n) time -
https://www.youtube.com/watch?v=1gxFE9EfanE&list=PLSIpQf0NbcCltzNFrOJkQ4J4AAjW3TSmA
I would really appreciate your comments and feedback on my video.
umangsomtiya8083 thankyou very much sir
Kanahaiya most welcome. 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.
karan316 How is this code working even when the numbers are very big numbers?
nitinkr432 Try in python.
def birthdayCakeCandles(ar): a=max(ar) return ar.count(a)
ryanfehr18 A small Java solution in O(n) time with O(1) space:
//Java 8 /* Initial Thoughts: We can keep a running max and update it if we find something larger, if we find something smaller we just keep looking and if we find something equal then we increment a counter variable Time Complexity: O(n) //We must check the height of every candle Space Complexity: O(1) //We only store a max and a frequency */ import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int tallest = 0; int frequency = 0; for(int i=0; i < n; i++){ int height = in.nextInt(); if(height > tallest){ tallest = height; frequency = 1; } else if(height == tallest) frequency++; } System.out.println(frequency); } }
You can find more HackerRank solutions like this here
Dsmeenakshi if height is less than tallest means...wat wil happen
ryanfehr18 As mentioned above:
/* if we find something smaller we just keep looking */
Basically, we don't care about candles < tallest
Dsmeenakshi thank you dude
wyrlvillazorda it's great..thank you!
randy_evered int max = 0; for (int v : a) if (v > max) max = v; int count = 0; for (int v : a) if (v >= max) count++; return count;sntz1 you can avoid doing two for loops by doing something along the lines of:
int max = 0; int count = 0; for (int v : a) { if (v > max) {max = v; count = 1;} else if (v == max) count++; } return count;
You can avoid iterating over the collection twice
srijanPaul ah yes, that's pretty smart.
SupreethMC Nice approch !
borisaltynnik Why not to use first array element like initial value of max ? Does it make any sense ?
static int birthdayCakeCandles(int n, int[] ar) { int max=ar[0]; int num=1; for (int i=1;i<n;i++){ if (ar[i]>max){ max=ar[i]; num=1; } else if (ar[i]==max) num++; } return num; }
iamkarthik49 Yeah ! I did the same thing but complicated it... Your's looks clean and clear thanks for the solution... :) .
santunu23 Nice one thanks for sharing,actually I have been working on this solution for a hour,finaly it work thanks dude.
Coder_Zuki No Probs Dude
renton284 This is not works, when a number in the array (last number of array) is bigger than then previous ones the result is not ok.
oliver_erhart The result would be 1 (correct). Because of the "else if" it doesn't get incremented again.
kvnfrederiksen0 HAH! I did the exact same thing. Nice.
int n = in.nextInt(); int j = 0; int max = 0; for(int i = 0; i < n; i++){ int x = in.nextInt(); if(x > max){ j = 1; max = x; } else if(x == max) j++; } System.out.print(j);
shantanu_bugadi simply wow.. what i got by this code is , try to do most of it simultaneously....
3001pratap static int birthdayCakeCandles(int n, int[] ar) { int count=0; Arrays.sort(ar); int tallest = ar[n-1]; for(int i=0; i<n; i++) { if(ar[i] == tallest) count++; } return count; }
fhrazib I think this can be improved a little bit using a break. We don't need to traverse all the array to get the total count when the array is sorted (here in ascending order) -
Arrays.sort(a); int tallestOne = a[n-1]; int count = 1; for(int i=n-2; i>=0; i--){ if(a[i]==tallestOne) { count++; } else break; }
hollybush Amazing solution mate!
zakiralig184 Cool!
pep1439 use
tallest = Integer.MIN_VALUE
instead, for general use.
stephen_kalletta very clean. I was sorting it and grabbing the last values, but I like this much better. thanks for sharing
Coder_Zuki Did This...! Seems Simple right?
static int birthdayCakeCandles(int n, int[] ar) { // Complete this function int size=ar.length; int tall=ar[0]; int count=0;
for(int i=0 ; i<size ; i++) { if(ar[i]>tall) { tall=ar[i]; } } for(int j=0;j<size;j++) { if(ar[j]==tall) count++; } return count; }
j_zambre66 [deleted]
nikhilgarakapat1 Give count=1;
j_zambre66 Can be done it like this.
int max = 0, count = 1; for (int v : ar) { if (v == max) count++; if (v > max) max = v; } return count;
disabajn Python:
return ar.count(max(ar))
hongsy YES.
#alltheothersolutionssuck
tom_wagner so good!!
antagonistpotato Sad to think I have discovered a one-liner just to see the discuissions and run into it here, but well, it happens
zimmah Ok, now THAT is short.
ATVex Counting the tallest candles.
n = int(input().strip()) height = [int(height_temp) for height_temp in input().strip().split(' ')] print(height.count(max(height)))
abhishek106 python is the best!
rudra_utpal Any problem with Java @abhishek106
import java.util.*; public class Solution { public static void main(String[] args) { Scanner scan=new Scanner(System.in); int n=scan.nextInt(); ArrayList<Integer> al=new ArrayList<Integer>(); for(int i=0;i<n;i++) al.add(scan.nextInt()); System.out.println(Collections.frequency(al,Collections.max(al,null))); } }
aaron_gomez1 While this works, you loop through the complete collection of values 3 times. First while reading it in the scanner. Second while scanning for the max value in the collection after reading it in and third to count the number of times it was found.
fernando_becerr1 anyway the time complexity is O(n+n+n) = O(n)
riaan_rvr I'm not too familiar with the BigO notation, so an honest question how does it become n+n+n shouldn't it be 3n rather as you are doing full loops over all elements?
hua_duy In BigOh notation, I believe they drop the constants in front of n because exponents matter so much more that the constants are negligible.
egrodo1 [deleted]egrodo1 Why do that instead of just?
n = int(raw_input().strip()) candles = map(int,raw_input().strip().split(' ')) print candles.count(max(candles))
randy_evered Wow, egrodo1, Python for the win!
zimmah javascript is just this:
const birthdayCakeCandles = a => (f => a.filter(v => v === f).length)(Math.max(...a));
Javascript isn't as hard as you think, you just need to understand it.
Modelmat I basically did the same thing!
pheepia7788 nice code! as I am a beginner, I think very long and messy code.... lol
GodsonLG Map object doesn't have the count attribute. You will get an error
Dulguun_Otgon [deleted]helderIO [deleted]
dandaraviteja n = int(input().strip()) ar = (list(map(int, input().strip().split(' ')))) print(ar.count(max(ar)))
paccel22 Thank you! I hadn't realized Python had a count method for lists that basically reduces the problem to 1 step.
Anubhav_singh what is the code of "count" function?? bcos when I am using loop instead of count fuction, its taking too much time to run. pls help??
ARXINO I am using loop but it is the same thing with ar.count(max(ar))
def birthdayCakeCandles(n, ar): max_height=0 double=0 for x in range(n): if ar[x]>max_height: max_height=ar[x] for y in range(n): if ar[y] == max_height: double+=1 return double
Shaurya_at_HR Thanks Arxino sir but after I started to run the code it showed a runtime error. I was asking that this function has 2 parameters but when below when the function is called it has only 1 parameter(ar). So what shall I do. Actually I am 13 years old and a beginner for coding in python. Pls help..
priyanshu__786 why did you used 2nd loop......i am not getting it can you please explain
sehan2 from collections import * def birthdayCakeCandles(n, ar): res = Counter(ar) return res.most_common()[0][1]
croqaz Exactly like I did! Nice!
rowlandoti This is not a correct solution but I do not know why it is passing. You should:
- Get the max element
- Return or print its count/frequency.
pWiszniewski In some test cases, the highest (max) element is the one most common - like in example case, where 3 is the most common element. It also passes some test cases. The correct solution is the one with Counter.
vinasatinfo just simple
return Counter(ar).most_common()[0][1]actually :-)
megha_sharma3333 Why is there a [0][1]?
Spiznak My solution finishd in about half the time, I would assume because I'm looping through it only once. Depends on whether you want a quick and dirty solution (mine) or one that's more Pythonic (in your case, can be more difficult to read):
n = input() arr = input().split() theMax, count = 0, 0 for x in arr: x = int(x) if x > theMax: theMax = x count = 1 elif x == theMax: count += 1 print(count)
icehongssii data=[10,3,333,333,333,0,13] tmp=data[0] #max data tmp2=0 #max data counter for i in range(len(data)-1): for j in range(i+1,len(data)): if(tmp<=data[j]): tmp=data[j] for k in range(len(data)): if(data[k]==tmp): tmp2+=1 print(tmp2)
this was my code. to get max_height, i use the loop twice. so it didn't work. but like you said, its working with single loop! thanks and it is so amazing that you used single looop... never imagined it
gb3010 I use max function in list. 7 out of 8 test cases work. I am not sure what's wrong with my code. Can you please have a look ?
n1=input().strip() # Input age value b1=input().strip().split() # Input list of candle heights def birthdayCakeCandles(n,b): c=0 z=max(b) for i in b: if i == z: c += 1 return(c) y=birthdayCakeCandles(n1,b1) print(y)
nicolasmontoya similar:
input() arr = list(map(int, input().split())) print(arr.count(max(arr)))
pWiszniewski EDIT: My mistake, i've missread the description :(. So the rest of the comment is invalid.
This doesn't work. It only counts how many times the highest element from the list (max(arr)) is in the list. In some test cases it works. In some it doesn't -> it's not a correct solution
brevnovak i don't understand. it's what we are looking for, is it not?? ;)
pWiszniewski You're right, I've missread the description.
wanwanzhang Hi I also use the same way to count in Python print(arr.count(max(arr))), but the result is always 2, none, which show me the worng answer. I do not know why it show me (2, none) instead of just 2.
sairajee Try to return the value from the function instead of printing.
return(ar.count(max(ar)))
EeXoR_Daniel_Ng python FTW:>
jeffmagill C# solution...
int tallest = height.Max(); int count = height.Count(c => c == tallest); Console.WriteLine(count);
indyHarcourt Or
int result = ar.Where(i => i == ar.Max()).Count();
unSatisfied That's O(n^2) since you're finding the max value on each iteration of the where statement. The best solution would be to use a single for/foreach loop, but jeffmagill's solution (looping through the list twice) is still much more efficient than this.
shatrudhankr [deleted]laysa_uchoa best comment
daeden I don't think you deserver a downvote, its solution i came up with the returned failed tests (tests timed out) for me.
I wish it was explained where other than comments why this wasn't a valid result.
leonardomiceli87 C# other way
var max = ar.Max(); return Array.FindAll(ar, s => s == max).Length;
Dalenguyen Nice to have modern JavaScript:
function birthdayCakeCandles(n, ar) { // Complete this function var max = Math.max(...ar); var result = ar.filter(c => c === max); return result.length; }
andrewseaton nice. I looped through the array to count occurances of "max". Using filter makes it so much simpler.
aJaxs 这个filter很机智
sameer_jain1 Exactly what I am thinking, but not sure why this question appearnign wrong at the moment, it doesnt have all the parameters so its throwing error to me
netris [deleted]netris You could even do this in a single line of code, like so:
function birthdayCakeCandles(ar) {
return ar.filter(i => i === Math.max(...ar)).length;
}But I found that this causes recursion issues. This works perfectly fine though:
function birthdayCakeCandles(ar) {
var max = Math.max(...ar);
return ar.filter(i => i === max).length;
}brecht_pynoo The first one is indeed overly complex, as it would search the max at each iteration
I used the reduce function instead, which allows for a single iteration, provided an extra local variable, like so
function birthdayCakeCandles(ar) { let max = 0; return ar.reduce( (count, i) => i === max ? count + 1 : i > max ? (max = i) && 1 : count, 0 ); }zimmah You can do it like this too:
const birthdayCakeCandles = a => (f => a.filter(v => v === f).length)(Math.max(...a));
cnps_bjjfan I did exactly what you did, but half the test failed and couldn't figure out why :P
zimmah Because it's horribly ineffiecient, causing it to time out (it takes more then 12 seconds so hackerrank cuts off the calculation to not waste processing power)
Because for every single item in the array it recalculates Math.max on the entire array, that's incredibly wastefull.
zimmah That oneliner kind of works, but it's extremely slow because it recalculates Math.max(ar) for every single item.
You can avoid that by storing the value of Math max like this:
const birthdayCakeCandles = a => (f => a.filter(v => v === f).length)(Math.max(...a));
That way it's still short, but also blazing fast. (Math max is implicitly stored as
fhere).
n_haque Oh filter and result.length is so much simpler! Mine was a little longer
function birthdayCakeCandles(ar) { let tallest = Math.max(...ar) let candleCount = 0 ar.forEach((candle) => candle === tallest ? candleCount++ : null) return candleCount }
careddu_m85 I used reduce to obtain the max, since spread operator can actually cause a stack overflow for large arrays.
josiah_mokobo function birthdayCakeCandles(ar) { return ar.filter(height => height === Math.max(...ar)).length }
josiah_mokobo You can filter out the max numbers from the array, and return its occurrance
bryank You can make it shorter with a bit of chaining.
function birthdayCakeCandles(ar) { const max = Math.max(...ar); return ar.filter(c => c === max).length; }
marinskiy Here is Python 3 solution from my HackerrankPractice repository:
n = int(input()) ar = list(map(int, input().split())) print(ar.count(max(ar)))
Feel free to ask if you have any questions :)
voodoorrj1990 [deleted]Tej_Pratap_Singh @marinskiy Are you still updating your repo from time to time?
Kanahaiya Hello friends,
Birthday cake candles hackerrank problem can be solved easily by using one for loop. The complexity of birthday cake candles hackerrank solution is O(n)
If interested to know more about the generic algorithm in details-
click here for the video explanation of generic algorithm with complexity analysis.
or you can click on the image too to follow youtube tutorial.
Here is the working solution:-
source code :
static int birthdayCakeCandles(int[] ar) { int maxCandleHeight = Integer.MIN_VALUE; int maxCandleFreqCount = 0; for (int i = 0; i < ar.length; i++) { if (ar[i] == maxCandleHeight) { maxCandleFreqCount++; } if (ar[i] > maxCandleHeight) { maxCandleHeight = ar[i]; maxCandleFreqCount = 1; } } return maxCandleFreqCount; }
Would really appreciate your feedback like, dislike , comment etc. on my video.
Do not forget to upvote, if you find it useful.
kaleeswaran35 Thanks for your insight Kanahaiya Gupta
Kanahaiya most welcome. if you dont mind.Would really appreciate your feedback like, dislike , comment etc. on my video.
leo_kamwathi HINT: If you get a timeout. Avoid using complex methods. Keep it simple.
soumith_kosaraju Didnt know .max() method was 'complex', thanks. Life = saved :)
zimmah It's not complex, you just need to store it. Instead of doing the same calcultion every time.
parkerbeck88 I didn't account for cases where my niece is turning 100,000.
lakshminarayan77 Problem with Test case 8 is there any thing diffrent in that test case??
NewSumanM [deleted]brian85 I tried to mimic the test case 8. My solution didn't work either and its total hack!
function birthdayCakeCandles(ar) { const isEdge = typeof parseInt(inputString[0], 10) === 'number' && inputString.length === 2 && inputString[1].split(' ').length > 0 ? true : false; const size = isEdge ? parseInt(inputString[0], 10) : ar.length; const tallest = ar.sort().reverse()[0]; const output = ar.reduce((acc, n) => { acc = n === tallest ? acc + 1 : acc; return acc; }, 0); console.log(output > size ? size: output) return output > size ? size : output; }I'm not sure what's going on either.
I tried this case:
200 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000It works and gives the answer 200 (there is 203 entries). however using the other one it crashes. The text file is too big to test in the editor.
[EDIT]
Figured it out
brian85 Issue was with Reversing the Array. Looks like it gets confused which array its revering. Wrapping in brackets was not enough. It seems like JS can't sort it correctly for some reason. I checked and every entry is type number so I'm not sure why .sort().reverse() doesn't work and why it needs to be in a bracket.
function birthdayCakeCandles(ar) { const size = ar.length; const candles = ar; const tallest = (candles.sort((a, b) => { return b - a; }))[0]; const output = candles.reduce((acc, n) => { acc = n === tallest ? acc + 1 : acc; return acc; }, 0); return output > size ? size : output; }juanitoyuco The question is how to proceed. The test cases were too large and causing a time out. Please help.
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