Prefix Sum and Sliding Window 01

Given an array A of n positive numbers. The task is to find the first Equilibrium 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.

Problem link :- click here

Code

// 	EQUILIBRIUM POINT

import java.util.Scanner;

public class Prefix_sum__Sliding_window_01
{
	
	static int equilibriumPoint(int[] arr, int n)
	{
		//	Calculating prefix sum array(psa)
		int[] psa = new int[n];
			
		psa[0] = arr[0];
		for(int i=1; i<n; i++)
			psa[i] = psa[i-1] + arr[i];
				
		/*
		NOTE: - 
			 if n==1 then the only element in the array is the equilibrium point
			 if n==2 then there will not be any equilibrium point
			 	
		To find the equilibrium point steps are: -
			 1) We have to iterate through each element starting from index 1,  `ref`
			 2) Then we have to compare the sum of elements to the left    <---- of `ref` 
			 		     	and the sum of elements to the right  ----> of `ref`
			 3) sum of elements to left of `ref` = 	psa[ref-1]
			 4) sum of elements to right of `ref` =	psa[n-1] - psa[ref]
			 		
		Since the given array contains only positive integers, there can be only 1 equilibrium point.
		*/
			
		if(n==1)
			return 1;
		if(n==2)
			return -1;
			
		for(int ref=1; ref<n; ref++)
		{
			int leftSum = psa[ref-1];
			int rightSum = psa[n-1] - psa[ref];
				
			if(leftSum == rightSum)
			{
				return (ref+1); 	//ref+1   for 1 index
			}
		}
		
		return -1;
	}
	
	public static void main(String[] args)
	{
		Scanner s = new Scanner(System.in);
		
		//Taking size of array and array as input from the user
		System.out.print("Enter the size of the array : ");
		int n = s.nextInt();
		
		int[] arr = new int[n];
		System.out.println("Enter the array elements : ");
		for(int i=0; i<n; i++)
			arr[i] = s.nextInt();
		
		
		int index = equilibriumPoint(arr, n);
		
		if(index == -1)
			System.out.println("\nEquilibrium Point doesn't exists");
		else
			System.out.println("\nEquilibrium Point is : " + index);
	}

}