Skip to content
Related Articles

Related Articles

Longest equilibrium Subarray of given Array

View Discussion
Improve Article
Save Article
  • Last Updated : 22 Aug, 2022
View Discussion
Improve Article
Save Article

Given an integer array arr of size N[], the task is to find the longest equilibrium subarray i.e. a subarray such that the prefix sum of the remaining array is the same as the suffix sum. 

Examples: 

Input: N = 3, arr[] =  {10, 20, 10}
Output: 1
Explanation: The longest subarray is {20}. The remaining prefix is {10} and suffix is {10}.
Therefore only one element in the subarray.

Input: N = 6, arr[] = {2, 1, 4, 2, 4, 1}
Output: 0
Explanation: The longest subarray is of size 0. 
The prefix is {2, 1, 4} and suffix is {2, 4, 1} and both has some 7.

 Input: N = 5, arr[] = {1, 2, 4, 8, 16}
Output: -1

 

Approach: This problem can be solved with two pointer technique based on the following idea:

Traverse from both the end. If the sum of prefix is less then increment the front pointer, otherwise, do the opposite. In this way we will get the minimum number of elements in prefix and suffix. So the subarray will have the maximum length because the remaining array has minimum elements.

Follow the steps mentioned below to implement the idea:

  • Let i be the left pointer initially at 0, and j be the right pointer initially at N-1.  
  • First, we will check if the sum of all elements is 0 or not. If 0 then the whole array will be the subarray.
  • Initialize two variable prefixSum = 0 and suffixSum = 0.
  • Traverse array from till i not equal to j.
    • If prefixSum <= suffixSum, then add arr[i] in prefixSum. And increment i by one.
    • Else check that if prefixSum > suffixSum, then add arr[i] in suffixSum. And decrement j by one.
    • Now check that if prefixSum equal to suffixSum then return the difference between i and j.
  • Otherwise, return -1.
     

 Below is the implementation  of the above approach:

C++




// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// C++ function to find out if prefixSum
// can be equal to the suffixSum
int isEqual(int n, int arr[])
{
    // Base Case
    if (n == 1) {
        return -1;
    }
     
    // Checking if the sum of all elements
    // is 0 or not. As the whole array can
    // also be a subarray.
    int sum=0;
    for(int i=0;i<n;i++)
    {
        sum+=arr[i];
    }
    if(sum==0)
    return n;
     
    int prefixSum = 0, suffixSum = 0;
    int i = 0, j = n - 1;
 
    // We are assuming that the ans is No
    int ans = -1;
 
    // Iterate till there is one element in the
    // array
    while (i <= j) {
 
        // If prefix sum is less or equal then
        // add it in the prefix sum and
        // increment i
        if (prefixSum <= suffixSum) {
            prefixSum += arr[i];
            i++;
        }
 
        // If suffix sum is less than add it in
        // the suffix sum and decrement j
        else if (prefixSum > suffixSum) {
            suffixSum += arr[j];
            j--;
        }
 
        // Check if both are equal or not? if they
        // are then make ans = "YES" and break
        if (prefixSum == suffixSum) {
            return j - i + 1;
        }
    }
    if (prefixSum == suffixSum)
        return 0;
 
    // return ans;
    return -1;
}
 
// Driver function
int main()
{
    int N = 3;
    int arr[] = { 10, 20, 10 };
 
    // Function call
    cout << isEqual(N, arr) << "\n";
    return 0;
}
 
// This code is contributed by Pushpesh Raj.


Java




// Java implementation for the above approach
import java.util.*;
public class GFG {
 
    // C# function to find out if prefixSum
    // can be equal to the suffixSum
    static int isEqual(int n, int[] arr)
    {
 
        // Base Case
        if (n == 1) {
            return -1;
        }
         
        // Checking if the sum of all elements
        // is 0 or not. As the whole array can
        // also be a subarray.
        int sum=0;
        for(int i=0;i<n;i++)
        {
            sum+=arr[i];
        }
        if(sum==0)
        return n;
 
        int prefixSum = 0, suffixSum = 0;
        int i = 0, j = n - 1;
 
        // We are assuming that the ans is No
        int ans = -1;
 
        // Iterate till there is one element in the
        // array
        while (i <= j) {
 
            // If prefix sum is less or equal then
            // add it in the prefix sum and
            // increment i
            if (prefixSum <= suffixSum) {
                prefixSum += arr[i];
                i++;
            }
 
            // If suffix sum is less than add it in
            // the suffix sum and decrement j
            else if (prefixSum > suffixSum) {
                suffixSum += arr[j];
                j--;
            }
 
            // Check if both are equal or not? if they
            // are then make ans = "YES" and break
            if (prefixSum == suffixSum) {
                return j - i + 1;
            }
        }
        if (prefixSum == suffixSum)
            return 0;
 
        // return ans;
        return -1;
    }
 
    // Driver function
    public static void main(String args[])
    {
        int N = 3;
        int[] arr = { 10, 20, 10 };
 
        // Function call
        System.out.println(isEqual(N, arr));
    }
}


Python3




# Python program for the above approach
def isEqual(n, arr):
 
    # Base Case
    if (n == 1):
        return -1
     
    # checking if the sum of all the elements
    # is 0 or not. As whole array can also
    # be the subarray
     
    Sum=0
    for i in range(n):
        Sum=Sum+arr[i]
    if(Sum==0):
        return n
 
    prefixSum,suffixSum = 0,0
    i,j = 0,n - 1
 
    # We are assuming that the ans is No
    ans = -1
 
    # Iterate till there is one element in the
    # array
    while (i <= j):
 
        # If prefix sum is less or equal then
        # add it in the prefix sum and
        # increment i
        if (prefixSum <= suffixSum):
            prefixSum += arr[i]
            i += 1
 
        # If suffix sum is less than add it in
        # the suffix sum and decrement j
        elif (prefixSum > suffixSum):
            suffixSum += arr[j]
            j -= 1
 
        # Check if both are equal or not? if they
        # are then make ans = "YES" and break
        if (prefixSum == suffixSum):
            return j - i + 1
 
    if (prefixSum == suffixSum):
        return 0
 
    # return ans;
    return -1
 
# Driver function
N = 3
arr = [10, 20, 10]
 
# Function call
print(isEqual(N, arr))


C#




// C# implementation for the above approach
using System;
class GFG {
 
// C# function to find out if prefixSum
// can be equal to the suffixSum
static int isEqual(int n, int[] arr)
{
     
    // Base Case
    if (n == 1) {
    return -1;
    }
     
    // Checking if the sum of all elements
    // is 0 or not. As the whole array can
    // also be a subarray.
    int sum=0;
    int k;
    for(k=0;k<n;k++)
     {
            sum+=arr[k];
     }
    if(sum==0)
    return n;
 
    int prefixSum = 0, suffixSum = 0;
    int i = 0, j = n - 1;
 
    // Iterate till there is one element in the
    // array
    while (i <= j) {
 
    // If prefix sum is less or equal then
    // add it in the prefix sum and
    // increment i
    if (prefixSum <= suffixSum) {
        prefixSum += arr[i];
        i++;
    }
 
    // If suffix sum is less than add it in
    // the suffix sum and decrement j
    else if (prefixSum > suffixSum) {
        suffixSum += arr[j];
        j--;
    }
 
    // Check if both are equal or not? if they
    // are then make ans = "YES" and break
    if (prefixSum == suffixSum) {
        return j - i + 1;
    }
    }
    if (prefixSum == suffixSum)
    return 0;
 
    // return ans;
    return -1;
}
 
// Driver function
public static void Main()
{
    int N = 3;
    int[] arr = { 10, 20, 10 };
 
    // Function call
    Console.WriteLine(isEqual(N, arr));
}
}


Javascript




<script>
        // JavaScript program for the above approach
        function isEqual(n, arr)
        {
         
            // Base Case
            if (n == 1) {
                return -1;
            }
             
            // Checking if the sum of all elements
            // is 0 or not. As the whole array can
            // also be a subarray.
            let sum=0;
            for(let i=0;i<n;i++)
            {
               sum+=arr[i];
            }
             if(sum==0)
              return n;
 
            let prefixSum = 0, suffixSum = 0;
            let i = 0, j = n - 1;
 
            // We are assuming that the ans is No
            let ans = -1;
 
            // Iterate till there is one element in the
            // array
            while (i <= j) {
 
                // If prefix sum is less or equal then
                // add it in the prefix sum and
                // increment i
                if (prefixSum <= suffixSum) {
                    prefixSum += arr[i];
                    i++;
                }
 
                // If suffix sum is less than add it in
                // the suffix sum and decrement j
                else if (prefixSum > suffixSum) {
                    suffixSum += arr[j];
                    j--;
                }
 
                // Check if both are equal or not? if they
                // are then make ans = "YES" and break
                if (prefixSum == suffixSum) {
                    return j - i + 1;
                }
            }
            if (prefixSum == suffixSum)
                return 0;
 
            // return ans;
            return -1;
        }
 
        // Driver function
        let N = 3;
        let arr = [10, 20, 10];
 
        // Function call
        document.write(isEqual(N, arr) + "<br>");
 
    </script>


Output

1

Time Complexity: O(N)
Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!