Equilibrium Point.cpp

// Equilibrium Point
// Difficulty: EasyAccuracy: 28.13%Submissions: 613K+Points: 2
// Given an array of integers arr[], the task is to find the first equilibrium point in the array.

// The equilibrium point in an array is an index (0-based indexing) such that the sum of all elements before that index is the same as the sum of elements after it. Return -1 if no such point exists. 

// Examples:

// Input: arr[] = [1, 2, 0, 3]
// Output: 2 
// Explanation: The sum of left of index 2 is 1 + 2 = 3 and sum on right of index 2 is 0 + 3 = 3.
// Input: arr[] = [1, 1, 1, 1]
// Output: -1
// Explanation: There is no equilibrium index in the array.
// Input: arr[] = [-7, 1, 5, 2, -4, 3, 0]
// Output: 3
// Explanation: The sum of left of index 3 is -7 + 1 + 5 = -1 and sum on right of index 3 is -4 + 3 + 0 = -1.
// Constraints:
// 3 <= arr.size() <= 106
// -109 <= arr[i] <= 109

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int findEquilibrium(vector<int> &arr) {
        // code here
        int n=arr.size();
        int totSum=accumulate(arr.begin(),arr.end(),0);
        int leftSum=0;
        for(int i=0;i<n;i++){
            totSum-=arr[i];
            if(totSum==leftSum){
                return i;
            }
            leftSum+=arr[i];
        }
        return -1;
    }