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For the test cases which contains a single element, make sure that you print YES, otherwise you will fail the Test #6. I think it is a huge mistake of the author, because it is not specified at all!
What do you mean by, they don't expect that.
c=in.nextInt();
lets you continue to the next test case normally.
If your code works on all the other test cases normally, then all you have to do is this :
[JavaCode]
if( arrSize ==1){
//you must add this line
int c = in.nextInt(); //currentInput.
System.out.println("YES");
continue; //Continue to the next test case.
}
because sum is to be tsken from left and right and if no such lements exists then obviously answer should be "no" as sum is not equal to that value from either side..
Technically, if you've coded things to the problem statement, you don't even need to consider this special case, although it could technically shortcut the rest of your logic.
Exactly. Not only that, but a general solution would handle negatives and 0, too (which isn't specified). A single element isn't a special case, in my thinking, but rather than case where the sum of the left and the sum of the right of the single element are equal (0). shrug
But look at the sample input
2
3
1 2 3
4
1 2 3 3
the result is
NO
YES
if the single element should be YES, the the result should be
YES
YES
NO
YES
YES
right?
Oops i think i misunderstood the test description.
For input
2 -> how many test case will coming
3 -> how many element in the first test case
1 2 3 -> the actual first test case
4 -> how many element in the second test case
1 2 3 4 -> the actual second test case
I thought the 2, 3 and 4 are also test cases :(
But now it's clear to me , thanks #warrior_of_light
// java method which passes 6 test cases, 7th test case fails. what is //the problem?
// Complete the balancedSums function below.
static String balancedSums(List<Integer> arr) {
int n = arr.size();
int i = 0;
int j = n-1;
int left = 0;
int right = 0;
while ( i < n && j >= 0)
{
if ( left == right && i == j) {
return "YES";
}
if ( left > right) {
right += arr.get(j);
j--;
} else
{
left += arr.get(i);
i++;
}
}
return "NO";
}
Before returning "NO", just check if left and right are equal or not, if they are equal return "YES". Then it will pass all the test cases. By the way your code is really good.
If it makes you feel better I tried it, and it has array index out of bound exception. Just revisit it and you'll see. And once again it's clear even without trying. You start at i=0 and you are trying to access item at i-1 which is index -1
I can see that the code passes the test cases, but you still have Undefined Behavior for accessing an out of bounds index. a[-1] where a is an array is always undefined behavior. It means anything can happen, including your program appearing to behave correctly.
No it works probably because the 4 bytes before is accessible and contains 0.
This is a classical undefined behavior case and a case of « it works on my computer syndroma ».
Running compiling/running this code with a different compiler (or on a different arch) may break. Just test it with address sanitizer (-fsanitize=address in gcc/clang) or a similar tool …
If it's in C/C++, there's no array bounds checking, so there would not be an exception. But as others have said, it's still not correct. It will read from the memory location just before the start of the array. If by chance or due to the specific implementation of the compiler, that memory location happens to hold 0, it will work. On a different system, or with a different compiler, or with different conditions, that memory location may have different contents that change the result.
u shud get yes as c element whose value is 5, sum of all elements to the left of it (suml) and sum of all elements on right of it(sumr) r equal:
suml = sumr
1+2+3+4 = 4+3+2+1
int i = 0;
int j = array.length - 1;
int sum = 0;
while (i != j) {
if(sum >= 0) {
sum -= array[j];
j--;
} else {
sum +=array[i];
i++;
}
}
return sum == 0 ? true : false;
Yep, I realized that too! I was looking to see if someone got that and you stated it more succinctly than I would have. It's much faster than the binary search method I originally thought of when I first read the problem.
It's not a mistake, He already said that if there are no element then take right/ left sum = 0. Now for single element there are no element on either side so right and left sum is zero which makes it a "YES"
If the test case contains a single element then there is no left and right sections so the answer should be 'NO'. Good post, I read this too late and feel a little cheated :)
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t;
cin>>t;
for(int i=0;i>n;
int a[200005];
for(int k=0;k>a[k];
}
int flag=0;
for(int k=0;k
You can keep two variables for left and right sum and set them to zero initially. That way, you don't have to write especially for the test case with 1 element.
its really embarrasing. my code passes against the first test and then on submission it passes against 0 , 1 ,2 , 5 while it gets timed out on 3 , 4 and fails on 6. i tried your hack as well but no success. can anybody point out whats wrong i am doing here.
public class SharlockArraySum {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (true) {
if (in.hasNextLine()) {
String s = in.nextLine();
if (s.contentEquals("")) {
break;
} else {
String[] strings = s.split(" ");
if (strings.length > 1) {
int[] numbers = new int[strings.length];
for (int i = 0; i < strings.length; i++) {
numbers[i] = Integer.parseInt(strings[i]);
}
checkIfLeftEqualToRight(numbers);
}else if(strings.length == 1){
}
private static void checkIfLeftEqualToRight(int[] integers) {
int totalIntArrayToLoop = integers.length - 1;
int sum = getTotalSum(integers, 0, totalIntArrayToLoop);
for (int i = 0; i <= totalIntArrayToLoop; i++) {
if (i == 0 || i == totalIntArrayToLoop) {
int remained = sum - integers[i];
if (remained == 0) {
System.out.println("YES");
return;
}
} else {
int toSubtract = getTotalSum(integers, 0, i - 1);
int resultToCheck = sum - (toSubtract * 2) - integers[i ];
if (resultToCheck == 0) {
System.out.println("YES");
return;
}
}
}
System.out.println("NO");
return;
}
private static int getTotalSum(int[] num, int start, int stop) {
int returnSum = 0;
for (int i = start; i <= stop; i++) {
returnSum = returnSum + num[i];
}
return returnSum;
}
I would suggest making the functions inline. You are passing an array to a function in checkIfLeftEqualToRight, which is as it is expensive, and on top of it the maximum array size is 10^5, adding all the more to the running time of the program. Tell me if it works.
the code has been written in java and sorry for my lack of knowledge but i never heard inline function in java. i tried to look up but didnt find much answere to that. apart from it from my understanding of java , we are passing the refrence of the array object but not the object itself so method is just extension of seprated code. i might be wrong.. in such case i would welcome updated information.
apart from it the logic that i have written it works in two 4 cases while fails in 3. in two timed out is there while in 1 it fails. i am trying to understand what am i doing wrong here logically.
I apologize for not being clear enough. I meant substitute the function call with the function itself. You might have to make slight modifications in the code, which you should be able to do with a little effort. In the else part of your checkifleftequaltoright, you are calling the gettotalsum function, and the call is within the for loop. I think that could be a cause of problem. Try substituting the function call with the function code and see if it works. Have you asked the experts while posting your question?
Tip: Always format your code while you ask a question. The greater the readability of your code, the more likely you get an answer. I have formatted it this time for you. I hope somebody else answers, if my response is not good enough.
public class SharlockArraySum
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
while (true)
{
if (in.hasNextLine())
{
String s = in.nextLine();
if (s.contentEquals(""))
{
break;
}
else
{
String[] strings = s.split(" ");
if (strings.length > 1)
{
int[] numbers = new int[strings.length];
for (int i = 0; i < strings.length; i++) {
numbers[i] = Integer.parseInt(strings[i]);
}
checkIfLeftEqualToRight(numbers);
}
else if(strings.length == 1)
{
// if(Integer.parseInt(strings[0]) == 1)
//{
// System.out.println("YES");
//}
}
}
}
else
{
break;
}
}
}
private static void checkIfLeftEqualToRight(int[] integers)
{
int totalIntArrayToLoop = integers.length - 1;
int sum = getTotalSum(integers, 0, totalIntArrayToLoop);
for (int i = 0; i <= totalIntArrayToLoop; i++)
{
if (i == 0 || i == totalIntArrayToLoop)
{
int remained = sum - integers[i];
if (remained == 0)
{
System.out.println("YES");
return;
}
}
else
{
int toSubtract = getTotalSum(integers, 0, i - 1);
int resultToCheck = sum - (toSubtract * 2) - integers[i];
if (resultToCheck == 0)
{
System.out.println("YES");
return;
}
}
}
System.out.println("NO");
return;
}
private static int getTotalSum(int[] num, int start, int stop)
{
int returnSum = 0;
for (int i = start; i <= stop; i++)
{
returnSum = returnSum + num[i];
}
return returnSum;
}
}
Here If you see n :
1<= n <= 10 ^ 5
So it is not right way to call getSum method everytime for given array which has lenght of 10 ^ 5.
Try not to call getSum method many times as it is too expensive.
Seel my code below if it can help you out:
import java.io.;
import java.util.;
public class Solution {
int n;
int[] nums;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t =sc.nextInt();
Solution[] s = new Solution[t];
for(int i = 0 ; i < t ; i++ )
{
s[i] = new Solution();
s[i].n = sc.nextInt();
s[i].nums = new int[s[i].n];
for(int j =0 ; j < s[i].n ;j++)
{
s[i].nums[j] = sc.nextInt();
}
boolean match = false;
int totalSum = getSum(0, s[i].n - 1, s[i].nums);
int previousSum = 0, leftSum =0, rightSum = 0;
for( int k = 0 ; k < s[i].n ; k++ )
{
previousSum = previousSum + s[i].nums[k];
leftSum = previousSum - s[i].nums[k];
rightSum = totalSum - leftSum - s[i].nums[k];
if(leftSum == rightSum)
{
match = true;
break;
}
}
if(match)
System.out.println("YES");
else
System.out.println("NO");
}
}
public static int getSum(int startIndex, int endIndex, int[] arr)
{
int sum =0;
for(int i = startIndex; i<= endIndex; i++)
sum = sum + arr[i];
return sum;
}
My logic is similar to yours. But I have declared everything as int. Large number inputs are failing in my case and i am pretty sure that my logic is correct. How can we handle large sums?
This is utter foolishness of the writer, for an array of size there exists no element to the right or left of the found element has hence the entire idea of test case, the only I was failing, is completely absurd.
I think that author has not made any mistakes here..It is properly mentioned in the constraints that
1<=N<=100000
and
also mentioned that ** If there are no elements to the left/right, then the sum is considered to be zero. **
Considering both the statements,above test case is a valid one.
I think that author has not made any mistakes here..It is properly mentioned in the constraints that
1<=N<=100000
and
also mentioned that ** If there are no elements to the left/right, then the sum is considered to be zero. **
Considering both the statements,above test case is a valid one.
Havealookatthis:)staticStringsolve(int[]a){// Complete this functionintsum=0;intlsum=0;intn=a.length;for(inti=0;i<n;i++){sum=sum+a[i];}intrsum=sum;for(inti=0;i<n;i++){rsum=rsum-a[i];if(rsum==lsum){return"YES";}lsum=lsum+a[i];}return"NO";}
It is not necessary to make a separate explicit condition that prints "YES" when the size of array is equal to one.
It all depends on how a person structures the code. So, this condition could be contained implicitly within the code as a part of one general condition for checking whether the two parts are equal.
And when the size of the array is equal to one then the variables of the global condition:
easy ... just add all the elements of the array!
in a loop as the array elements to a new variable and substract the upcoming elemet from the total!
compare the two! if equal print yes else no!
passes
1
1
1
@himadri77 it is not necessary to put a manual condition for one element in the array, i think your logic is wrong rather than the author. Here is my code and its working perfectly...
def sherlock_algo(arr):
l = 0
u = len(arr)
index = (l+u)//2
while l<=u :
sum1 = sum(arr[0:index])
sum2 = sum(arr[index+1:])
if sum1 == sum2:
return index
elif sum1 > sum2:
u = index-1
else:
l = index + 1
index = (l + u) // 2
return -1
t = int(input()) #testcases
for i in range(t):
n = int(input()) # length of list
list1 = list(int(x) for x in input().split())
result = sherlock_algo(list1)
if result == -1:
print("NO")
else:
print("YES")
Sherlock and Array
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For the test cases which contains a single element, make sure that you print YES, otherwise you will fail the Test #6. I think it is a huge mistake of the author, because it is not specified at all!
For example for input:
111Make sure your answer is:
YES+1, thanks!
https://www.hackerrank.com/challenges/sherlock-and-array/submissions/code/105519037
putting condition is not necessary.check this out.
int sum = accumulate(arr.begin(), arr.end(), 0); int count = arr[0]; int len = arr.size();
kudoos to you sir.......i was totally blank on the approach and you did it
I wasted hackos before reading your post. :(
+1 Thanks.
NOOO, NOT THE HAACKOS! :'-(
same here
Won't that cause other tests to fail?
edit: I just tried printing "YES" if input is length 1, it caused all tests to fail. Is there an actual workaround for this?
You need to do this:
if( arrSize ==1){
int c = in.nextInt(); System.out.println("YES");
continue; }
where c is the current input from STDN. and you continue so that you can go to the next testcase.
When I print YES if length is 1, then all the other test cases fail because they don't expect that.
What do you mean by, they don't expect that. c=in.nextInt(); lets you continue to the next test case normally. If your code works on all the other test cases normally, then all you have to do is this :
[JavaCode]
ya its working..but didn't understand
I checked against a custom input of 1 1 1
the answer comes out as YES. but why it is failing? need help
because sum is to be tsken from left and right and if no such lements exists then obviously answer should be "no" as sum is not equal to that value from either side..
same problem with me
And I wasted 6 months in this . Thank you :) . Finally I can be in peace . +1 :)
True they should rephrase it.
Technically, if you've coded things to the problem statement, you don't even need to consider this special case, although it could technically shortcut the rest of your logic.
Exactly. Not only that, but a general solution would handle negatives and 0, too (which isn't specified). A single element isn't a special case, in my thinking, but rather than case where the sum of the left and the sum of the right of the single element are equal (0). shrug
But look at the sample input
2
3
1 2 3
4
1 2 3 3
the result is
NO
YES
if the single element should be YES, the the result should be
YES
YES
NO
YES
YES
right?
If the given input n (here 3 and 4) is equal to 1, then only we have to print yes.
Oops i think i misunderstood the test description.
For input
2 -> how many test case will coming
3 -> how many element in the first test case
1 2 3 -> the actual first test case
4 -> how many element in the second test case
1 2 3 4 -> the actual second test case
I thought the 2, 3 and 4 are also test cases :(
But now it's clear to me , thanks #warrior_of_light
If n == 1, then definitely answer is YES, for n > 1, you have to check.
// java method which passes 6 test cases, 7th test case fails. what is //the problem? // Complete the balancedSums function below.
Before returning "NO", just check if left and right are equal or not, if they are equal return "YES". Then it will pass all the test cases. By the way your code is really good.
Hi,
Here is the video explanation of my solution with different approach in O(n) time-
https://youtu.be/xPHJH8RL3mo
any feedback or comment would be highly appreciated.
Didn't need to do it.
for( int i = 0 ; i < n ; i++)
for( int i = 0 ; i < n ; i++)
{
It's pretty obvious you have an ArrayIndexOutOfBoundsException even without trying
If it makes you feel better I tried it, and it has array index out of bound exception. Just revisit it and you'll see. And once again it's clear even without trying. You start at i=0 and you are trying to access item at i-1 which is index -1
Hi Buddy,
You have an array index out of bounds in u r code at line 24 because of negative index. Also, add a condition that if length is 1 display YES.
for( int i = 0 ; i < n ; i++) { lc += a[i-1];
Cheers. Good Day.
Hi Buddy,
You have an array index out of bounds in u r code at line 24 because of negative index. Also, add a condition that if length is 1 display YES.
for( int i = 0 ; i < n ; i++) { lc += a[i-1];
Cheers. Good Day.
I can see that the code passes the test cases, but you still have Undefined Behavior for accessing an out of bounds index. a[-1] where a is an array is always undefined behavior. It means anything can happen, including your program appearing to behave correctly.
Yes, the code is working without a crash because there is a statment to break the loop if condition match.
No it works probably because the 4 bytes before is accessible and contains 0.
This is a classical undefined behavior case and a case of « it works on my computer syndroma ».
Running compiling/running this code with a different compiler (or on a different arch) may break. Just test it with address sanitizer (-fsanitize=address in gcc/clang) or a similar tool …
you will get an exception
If it's in C/C++, there's no array bounds checking, so there would not be an exception. But as others have said, it's still not correct. It will read from the memory location just before the start of the array. If by chance or due to the specific implementation of the compiler, that memory location happens to hold 0, it will work. On a different system, or with a different compiler, or with different conditions, that memory location may have different contents that change the result.
There may not be an exception. It can still seg fault.
It's wrong in any case.
try like this lc += (i==0) ? 0 : a[i-1];
try to use
for( int i = 1 ; i < n ; i++) { lc += a[i-1]; rc -= a[i-1]; if(lc == rc-a[i]) { flag = 0; break; }
then you have to special case the array of length 1 that doesn't have an index at 1.
That's some great implementation. used the same logic but my code is way longer than yours
A minor change as follows will fix your minor problem:
Cheers
bro could u explain me what the que is plz? what if i have like 1 2 3 4 5 4 3 2 1 ....what answer should i get?
u shud get yes as c element whose value is 5, sum of all elements to the left of it (suml) and sum of all elements on right of it(sumr) r equal: suml = sumr 1+2+3+4 = 4+3+2+1
excellent!
From my point of view everithing pretty obvious imagine that one element it's a middle of array sum on left side = 0 as well as on right side = 0.
for the last line, you don't need to use ternary operator if all you want is to return a boolean
will be suffice
Yep, I realized that too! I was looking to see if someone got that and you stated it more succinctly than I would have. It's much faster than the binary search method I originally thought of when I first read the problem.
Great and Simple Solution!
fantastic algorithm
Awesome , Salute..
This fails following: n=4: 0 0 2 0
It passes all test cases, including n=4: 0 0 2 0 !!!! Just copy and paste the whole code, and do not forget to import java.util.Scanner !!!
You are right This is not also not work for 0 2 0 0
it fails for test input 0,0,2,0.
CHEERS
Only came to see the comments after realizing that...could have saved 10min...
It's not a mistake, He already said that if there are no element then take right/ left sum = 0. Now for single element there are no element on either side so right and left sum is zero which makes it a "YES"
Not really. I agree with the author. If you remove the single element, then 0 == 0. Thanks for the comment though. was helpfull.
If the test case contains a single element then there is no left and right sections so the answer should be 'NO'. Good post, I read this too late and feel a little cheated :)
+1 Thanks
hi.. failinf test case 3 and 4 due to time out.
me too same problem..do u resolve your problm.
include
include
include
include
include
using namespace std;
int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t; cin>>t; for(int i=0;i>n; int a[200005]; for(int k=0;k>a[k]; } int flag=0; for(int k=0;k
}
This is my code... Failing test case 3 and 4 due to time out.Kindly help.
yours code complexity is o(n^2).It can be solved in o(n).
how?
For future readers:
You can keep two variables for left and right sum and set them to zero initially. That way, you don't have to write especially for the test case with 1 element.
I failed at that test case too. Thank you!
its really embarrasing. my code passes against the first test and then on submission it passes against 0 , 1 ,2 , 5 while it gets timed out on 3 , 4 and fails on 6. i tried your hack as well but no success. can anybody point out whats wrong i am doing here.
public class SharlockArraySum {// if(Integer.parseInt(strings[0]) == 1){ // System.out.println("YES"); // } } } } else { break; } }
}
I would suggest making the functions inline. You are passing an array to a function in checkIfLeftEqualToRight, which is as it is expensive, and on top of it the maximum array size is 10^5, adding all the more to the running time of the program. Tell me if it works.
Hi , Thanks for the reply.
the code has been written in java and sorry for my lack of knowledge but i never heard inline function in java. i tried to look up but didnt find much answere to that. apart from it from my understanding of java , we are passing the refrence of the array object but not the object itself so method is just extension of seprated code. i might be wrong.. in such case i would welcome updated information.
apart from it the logic that i have written it works in two 4 cases while fails in 3. in two timed out is there while in 1 it fails. i am trying to understand what am i doing wrong here logically.
I apologize for not being clear enough. I meant substitute the function call with the function itself. You might have to make slight modifications in the code, which you should be able to do with a little effort. In the else part of your checkifleftequaltoright, you are calling the gettotalsum function, and the call is within the for loop. I think that could be a cause of problem. Try substituting the function call with the function code and see if it works. Have you asked the experts while posting your question?
Tip: Always format your code while you ask a question. The greater the readability of your code, the more likely you get an answer. I have formatted it this time for you. I hope somebody else answers, if my response is not good enough.
Please accept my thanks for answering to my queries as well as helping me out :)
You're welcome. Is your solution now passing all test cases?
unfortunately no , the issue is with logic i think. i am solving other puzzles in the mean while and letting my mind cool down for this logic :).
Here If you see n : 1<= n <= 10 ^ 5 So it is not right way to call getSum method everytime for given array which has lenght of 10 ^ 5. Try not to call getSum method many times as it is too expensive. Seel my code below if it can help you out: import java.io.; import java.util.;
public class Solution { int n; int[] nums;
}
Thanks a lot himadri77 :D
Thanx a lot..!! Saved my day :)
I Don't know how. But without checking for array length, my code passed all tests. Following is what i did-
My logic is similar to yours. But I have declared everything as int. Large number inputs are failing in my case and i am pretty sure that my logic is correct. How can we handle large sums?
Thanks @Tomsaju :) This approach really helped me passing the testcases #3 and #4 as i was getting timeout earlier.
Hey it's mentioned in the problem, if you do not find any element to its left or right. It should be considered as zero.
In this case for 1 left = 0 right = 0
both are same, print Yes
I think when he mentioned that in case of no element is present left/right, he did kind of gave a hint about such a scenario.
This is utter foolishness of the writer, for an array of size there exists no element to the right or left of the found element has hence the entire idea of test case, the only I was failing, is completely absurd.
I think that author has not made any mistakes here..It is properly mentioned in the constraints that 1<=N<=100000 and also mentioned that ** If there are no elements to the left/right, then the sum is considered to be zero. ** Considering both the statements,above test case is a valid one.
I think that author has not made any mistakes here..It is properly mentioned in the constraints that 1<=N<=100000 and also mentioned that ** If there are no elements to the left/right, then the sum is considered to be zero. ** Considering both the statements,above test case is a valid one.
what if there are only two elements in the array?
In that case answer would be NO no matter what the number is.
Example
4 5
if i=0 leftsum = 0 and rightsum = 5 if i=1 leftsum = 4 and rightsum = 0 (also note that Ai>=1)
If that is correct, the test 0 for:
2
3
1 2 3
4
1 2 3 3
should'nt be this?
YES
YES
NO
YES
YES
the output expected is:
NO
YES
I assumed that is for the row 3 and 5
you saved my day, thanks
+1, thanks!
I wasted hours thinking why Test #6 is failing before coming to this discussion forum and seeing your comments.
Thanks..
+1. This case #6 is wrong.
thanks for your support , although we have atmost same approach
How to solve the problem "Terminated due to timeout"?
No its not mistake by writer, there is writen " If there are no elements to the left/right, then the sum is considered to be zero."
consider
** happy coding**
It is not necessary to make a separate explicit condition that prints "YES" when the size of array is equal to one. It all depends on how a person structures the code. So, this condition could be contained implicitly within the code as a part of one general condition for checking whether the two parts are equal. And when the size of the array is equal to one then the variables of the global condition:
will have values:
Here is my solution for Java:
easy ... just add all the elements of the array! in a loop as the array elements to a new variable and substract the upcoming elemet from the total! compare the two! if equal print yes else no! passes 1 1 1
6th test case
There is no need to have a separate check for singleton lists.
How did you thought such an Amazing idea. :)
Hi ,
I have uploaded a video tutorial which can help you to how he came up with this idea.
https://youtu.be/52R2pLDjUBw
Guys do visit his channel once there is no harm that. The beauty is that he has soved question in very good way and they are easier to underastand.
Hi All,
I have uploaded a video tutorial on Sherlock and Array solution.
https://youtu.be/52R2pLDjUBw
Please have a look and suggest if you have better apporach to solve this problem.
Regards,
Kanahaiya Gupta
its not necessary that you will necessarily fail.depends on code. mine worked without it
It mentioned in the question
C++ solution :
@himadri77 it is not necessary to put a manual condition for one element in the array, i think your logic is wrong rather than the author. Here is my code and its working perfectly...
def sherlock_algo(arr): l = 0 u = len(arr) index = (l+u)//2 while l<=u : sum1 = sum(arr[0:index]) sum2 = sum(arr[index+1:])
t = int(input()) #testcases for i in range(t):
This won't need any condition checking