In this Sherlock and Array HackerRank solution, Watson gives Sherlock an array of integers. His challenge is to find an element of the array such that the sum of all elements to the left is equal to the sum of all elements to the right.
Example
is between two subarrays that sum to .
The answer is since left and right sum to .
You will be given arrays of integers and must determine whether there is an element that meets the criterion. If there is, return YES. Otherwise, return NO.
Function Description
Complete the balancedSums function in the editor below.
balancedSums has the following parameter(s):
- int arr[n]: an array of integers
Returns
- string: either
YESorNO
Input Format
The first line contains , the number of test cases.
The next pairs of lines each represent a test case.
– The first line contains , the number of elements in the array .
– The second line contains space-separated integers where .
Constraints
Sample Input
2
3
1 2 3
4
1 2 3 3
Sample Output
NO
YES
Explanation
For the first test case, no such index exists.
For the second test case, , therefore index satisfies the given conditions.
Sherlock and Array HackerRank Solution
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Problem Solution in C#
public static string balancedSums(List<int> arr)
{
{
int length = arr.Count();
int left = 0;
int right = 0;
int last = 0;
for (int i=0;i<length;i++)
{
left += arr[i];
}
for (int i=length-1;i>-1;i--)
{
left-=arr[i];
right+=last;
if (left==right)
{
return "YES";
}
last = arr[i];
}
return "NO";
}
}Problem Solution in Java
public static String balancedSums(List<Integer> arr) {
if (arr.size() == 1 ){ return "YES"; }
int left = 0, sumLeft = arr.get(left), right = arr.size() - 1 , sumRight = arr.get(right);
while ( left < right ){
if (sumLeft < sumRight ){
sumLeft += arr.get(++left);
} else if (sumLeft > sumRight ){
sumRight += arr.get(--right);
} else if (sumLeft == sumRight) {
sumLeft += arr.get(++left);
sumRight += arr.get(--right);
}
}
return ( sumLeft == sumRight ) ? "YES" : "NO";
}
}Problem Solution in JavaScript
function balancedSums(arr) {
let sumLeft = 0;
let sumRigth = arr.reduce((acc, cur) => acc + cur);
for(let i = 0; i < arr.length; i++) {
sumLeft += arr[i - 1] || 0;
sumRigth -= arr[i];
if(sumRigth === sumLeft) return "YES";
}
return "NO"
}Problem Solution in Python
def balancedSums(arr):
# Write your code here
a=sum(arr)
add=0
for x in arr:
if a-x==add:
return "YES"
elif add>a:
return "NO"
add+=x
a-=x
return "NO"