Prefix Sum and Sliding Window 06

Given two binary arrays arr1[] and arr2[] of same size N. Find length of the longest common span [i, j] where j>=i such that arr1[i] + arr1[i+1] + .... + arr1[j] = arr2[i] + arr2[i+1] + .... + arr2[j].

Problem link :- click here

Code

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

//	LONGEST SPAN IN TWO BINARY ARRAYS

public class Prefix_sum__Sliding_window_06
{

	static int longestSpanInTwoBinaryArrays(int arr1[], int arr2[], int n)
	{
		int arr[] = new int[n];
		
		//	Initializing the array `arr`
		for(int i=0; i<n; i++)
			arr[i] = arr1[i] - arr2[i];
		
		/*
		 	METHOD: -
		 		arr1 = {0, 1, 0, 0, 0, 0}
		 		arr2 = {1, 0, 1, 0, 0, 1}
		 		
		 		arr   =  arr1 - arr2
		 		arr   =  {-1, 1, -1, 0, 0, -1}
		 		
		 		psa   =  {-1, 0, -1, -1, -1, -2}
		 		
		 		possible lengths: -
		 			i) When sum == 0
		 				maxLen = i + 1 
		 						= 1 + 1 
		 						= 2
		 			ii) When hm.containsKey(sum)
		 				maxLen = i - hm.get(sum)
		 				
		 					a)	= 2 - 0
		 						= 2
		 					
		 					b)	= 3 - 0
		 						= 3
		 						
		 					c)	= 4 - 0
		 						= 4
		 	
		 		Among all these possible lengths, the longest is 4.
		 */
		
		Map<Integer, Integer> hm = new HashMap<>();
		int maxLen = 0;
		
		int sum = 0;
		for(int i=0; i<n; i++)
		{
			sum += arr[i];
			
			if(sum == 0)
				maxLen = i + 1;
			
			if( hm.containsKey(sum) )
				maxLen = Math.max(maxLen, i - hm.get(sum) );
			else
				hm.put(sum, i);
		}
		
		return maxLen;
	}
	
	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[] arr1 = new int[n];
		System.out.println("Enter the array elements : ");
		for(int i=0; i<n; i++)
			arr1[i] = s.nextInt();
		
		int[] arr2 = new int[n];
		System.out.println("Enter the array elements : ");
		for(int i=0; i<n; i++)
			arr2[i] = s.nextInt();
		
		int maxLen = longestSpanInTwoBinaryArrays(arr1, arr2, n);
		System.out.println(maxLen);
	}

}