EquilibriumPoint-GFGQ.cpp

/*
Equilibrium Point

Given an array A of n positive numbers. The task is to find the first Equilibium Point in the array. 
Equilibrium Point in an array is a position such that the sum of elements before it is equal to the sum of elements after it.

Note: Retun the index of Equilibrium point. (1-based index)

Example 1:

Input: 
n = 5 
A[] = {1,3,5,2,2} 
Output: 3 
Explanation:  
equilibrium point is at position 3 
as elements before it (1+3) = 
elements after it (2+2). 
 

Example 2:

Input:
n = 1
A[] = {1}
Output: 1
Explanation:
Since its the only element hence
its the only equilibrium point.
 

Your Task:
The task is to complete the function equilibriumPoint() which takes the array and n as input parameters and returns the point of equilibrium. Return -1 if no such point exists.

Expected Time Complexity: O(n)
Expected Auxiliary Space: O(1)

Constraints:
1 <= n <= 106
1 <= A[i] <= 108
*/

//{ Driver Code Starts
#include <iostream>
using namespace std;


// } Driver Code Ends
class Solution{
    public:
    // Function to find equilibrium point in the array.
    // a: input array
    // n: size of array
    int equilibriumPoint(long long a[], int n) {
    
        // Your code here
        int pre = 0, post = 0;
	    for (int i = 0; i < n; i++)	post += a[i];
	    for (int i = 0; i < n; i++) {
		    post -= a[i];
		    if (pre == post) return i+1;
		    pre += a[i];
	    }
	    return -1;
    }

};

//{ Driver Code Starts.


int main() {

    long long t;
    
    //taking testcases
    cin >> t;

    while (t--) {
        long long n;
        
        //taking input n
        cin >> n;
        long long a[n];

        //adding elements to the array
        for (long long i = 0; i < n; i++) {
            cin >> a[i];
        }
        
        Solution ob;

        //calling equilibriumPoint() function
        cout << ob.equilibriumPoint(a, n) << endl;
    }
    return 0;
}

// } Driver Code Ends