class Solution {
    // Function to find equilibrium point in the array.
    public static int equilibriumPoint(long a[], int n) {

        //We store the sum of all array elements.
        long sum = 0;
        for (int i = 0; i < n; i++) 
           sum += a[i];

        //sum2 is used to store prefix sum.
        long sum2 = 0;
        int ans = -1;

        for (int i = 0; i < n; i++) {
            
            //Leaving out the value of current element from suffix sum.
            sum = sum - a[i];
            
            //Checking if suffix and prefix sums are same.
            if (sum2 == sum) {
                //returning the index or equilibrium point.
                return (i + 1);
            }
            
            //Adding the value of current element to prefix sum.
            sum2 = sum2 + a[i];
        }
        return -1;
    }
}

// Efficient Method : Time Complexity : O(n), Auxilliary space : O(1)

import java.util.*;
import java.io.*;

class GFG 
{ 
    static boolean checkEquilibrium(int arr[], int n)
    {
    	int sum = 0;

    	for(int i = 0; i < n; i++)
    	{
    		sum += arr[i];
    	}

    	int l_sum = 0;

    	for(int i = 0; i < n; i++)
    	{
    		if(l_sum == sum - arr[i])
    			return true;

    		l_sum += arr[i];

    		sum -= arr[i];
    	}

    	return false;
    }

    public static void main(String args[]) 
    { 
       int arr[] = {3, 4, 8, -9, 20, 6}, n = 6;

       System.out.println(checkEquilibrium(arr, n));
    } 
}




// Naive Method : Time Complexity : O(n^2)

    static boolean checkEquilibrium(int arr[], int n)
    {
    	for(int i  = 0; i < n; i++)
    	{
    		int l_sum = 0, r_sum = 0;

    		for(int j = 0; j < i; j++)
    			l_sum += arr[j];

    		for(int j = i + 1; j < n; j++)
    			r_sum += arr[j];

    		if(l_sum == r_sum)
    			return true;
    	}

    	return false;
    }