Last Updated: 18 Nov, 2020
Check if sum of left side of array equal to right side of array
Easy
You are given an array Arr consisting of n integers. You need to find the equilibrium index of the array. An index is considered as equilibrium index if the sum of elements to the left of that index is equal to the sum of elements to the right of it.
Note:
The array follows 0-based indexing, so you need to return the 0- based index of the element. If there are multiple indices which satisfy the given condition, then return the left-most index. If no such index is present in the array, return -1.
Input Format:
The first line of the input contains an integer T, denoting the number of test cases.
The first line of each test case contains the integer N, denoting the size of the array.
The second line of each test case contains N space-separated integers denoting the array elements.
Output Format:
The only line of output of each test case contains a single integer which corresponds to the equilibrium index for the array.
#### Note : You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints:
1 <= T <= 50
1 <= N <= 10^5
0 <= Arr[i] <= 10^4
Time Limit: 1 sec
Approaches
- We can try the brute force approach in which for every index, we can run a loop to the left of that index and to the right of that index, and check if the sum of elements is equal or not.
- Let us initialise the ans = -1, which will contain our valid index.
- Since index 0 and n - 1 cannot be possible candidates for the answer, we can start the loop from i = 1 to i = n - 2.
- Now for every i, let leftSum = 0 and rightSum = 0.
- For j = 0 to j = i - 1, do leftSum = leftSum + arr[j].
- For k = i + 1 to k = n - 1, do rightSum = rightSum + arr[k].
- If leftSum = rightSum, then return i.
- After the loop of i ends, return -1, since we have not found a valid index.
- Since we don’t want to calculate the sum of elements, again and again, we can store the sum of the elements in the form of prefix sum. Let prefSum[] be the array which stores the prefix sum.
- So if we are at index i, the sum of elements to left of index i is stored at leftSum = prefSum[i - 1], given that i-1 is a valid array index.
- Let arraySum denote the sum of the complete array.
- Now to find the sum of elements to the right of the index i, rightSum = total sum - (sum of elements to the left of i )+ arr[i].
- The above statement can be simplified to rightSum = arraySum - prefSum[i - 1] + arr[i]. Moreover, rightSum = arraySum - prefSum[i] (since prefSum[i - 1] + arr[i] = prefSum[i]).
- If we get leftSum = rightSum at any index, return that index.
- After for loop ends return -1, since there is no valid index.
- We can even skip the prefix sum array and replace it with some variables.
- Let arraySum be the sum of the array and leftSum = 0, which will denote the sum of elements to the left of index i.
- Now for i = 0 to i = n - 1, as we move from left to right do leftSum = leftSum + arr[i].
- The sum of elements to the right of i can be found as rightSum = arraySum - leftSum - arr[i].
- If we have leftSum = rightSum, then return i.
- After the loop is ended, return -1, since no such index exists.