23. Equilibrium Point

The problem can be found at the following link: Question Link

My Approach

Petty easy question.

• To find the equilibrium point, I calculate the total sum of the elements in the array. Then,

• I iterate through the array and maintain a prefix sum preSum and subtract the current element from the total sum.

  • If the remaining sum == preSum equals, I return the current index as the equilibrium point.

Time and Auxiliary Space Complexity

• Time Complexity: O(N) where N is the number of elements in the array. We iterate through the array twice: once to calculate the total sum and once to find the equilibrium point.

• Auxiliary Space Complexity: O(1). We only use a constant amount of extra space to store the sum and preSum.

Code (C++)

class Solution {
public:
    int equilibriumPoint(long long a[], int n) {
        long long sum = 0, preSum = 0;
        for(int i = 0; i < n; ++i)
            sum += a[i];
            
        for(int i = 0; i < n; ++i){
            sum -= a[i];
            if(sum == preSum)
                return i+1;
            preSum += a[i];
        }
        return -1;
    }
};

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