Equilibrium Index

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Problem statement

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 an equilibrium index if the sum of elements of the array to the left of that index is equal to the sum of elements to the right of it.

Note:

1. The array follows 0-based indexing, so you need to return the 0-based index of the element.
2. Note that the element at the equilibrium index won’t be considered for either left sum or right sum.
3. If there are multiple indices which satisfy the given condition, then return the left-most index i.e if there are indices i,j,k…. which are equilibrium indices, return the minimum among them
4. If no such index is present in the array, return -1.

Detailed explanation ( Input/output format, Notes, Images )

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
-10^4 <= Arr[i] <= 10^4
Time Limit: 1 sec

Sample Input 1:

1
6
1 7 3 6 5 6

Sample Output 1:

Explanation for Sample Input 1:

The sum of elements to the left of arr[3] = 1 + 7 + 3 = 11.
The sum of elements to the right of arr[3] = 5 + 6 = 11.
Hence the answer is 3.

Sample Input 2:

2
6
1 2 2 9 3 2
4
1 2 3 4

Sample Output 2:

3
-1

Explanation for Sample Input 2:

In the first test case of sample input 2, the sum of elements on the left of index 3 (arr[3] = 9) is five which is equal to the sum of elements on the right of arr[3].
For the second test case, since no such index exists, we print -1.

Hint 1

Hint 2

Hint 3

Naively find the answer for every index.

Approaches (3)

Approach 1

Approach 2

Approach 3

Brute Force

We can try the brute force approach in which we consider every index starting from 0 as a possible candidate for equilibrium index, and to check if it is really an equilibrium index, we can calculate the sum of elements on both sides of that index, and check if they are equal or not. 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.

Time Complexity

O(N^2), where N is the size of the array.

Space Complexity

O(1), as we are using constant extra memory.

(100% EXP penalty)

Equilibrium Index

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