Skip to content
Related Articles

Related Articles

Improve Article

Count subarrays having sum of elements at even and odd positions equal

  • Difficulty Level : Hard
  • Last Updated : 22 Apr, 2021

Given an array arr[] of integers, the task is to find the total count of subarrays such that the sum of elements at even position and sum of elements at the odd positions are equal.

Examples:

Input: arr[] = {1, 2, 3, 4, 1}
Output: 1
Explanation: 
{3, 4, 1} is the only subarray in which sum of elements at even position {3, 1} = sum of element at odd position {4}

Input: arr[] = {2, 4, 6, 4, 2}
Output: 2
Explanation: 
There are two subarrays {2, 4, 6, 4} and {4, 6, 4, 2}.

 

Approach: The idea is to generate all possible subarrays. For each subarray formed find the sum of the elements at even index and subtract the elements at odd index. If the sum is 0, count this subarray else check for the next subarray.



Below is the implementation of the above approach:

C++




// C program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
void countSubarrays(int arr[], int n)
{
     
    // Initialize variables
    int count = 0;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        int sum = 0;
 
        for(int j = i; j < n; j++)
        {
             
            // Check if position is
            // even then add to sum
            // then add it to sum
            if ((j - i) % 2 == 0)
                sum += arr[j];
 
            // Else subtract it to sum
            else
                sum -= arr[j];
 
            // Increment the count
            // if the sum equals 0
            if (sum == 0)
                count++;
        }
    }
 
    // Print the count of subarrays
    cout << " " << count ;
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 2, 4, 6, 4, 2 };
 
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    countSubarrays(arr, n);
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C




// C program for the above approach
#include <stdio.h>
 
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
void countSubarrays(int arr[], int n)
{
     
    // Initialize variables
    int count = 0;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        int sum = 0;
 
        for(int j = i; j < n; j++)
        {
             
            // Check if position is
            // even then add to sum
            // then add it to sum
            if ((j - i) % 2 == 0)
                sum += arr[j];
 
            // Else subtract it to sum
            else
                sum -= arr[j];
 
            // Increment the count
            // if the sum equals 0
            if (sum == 0)
                count++;
        }
    }
 
    // Print the count of subarrays
    printf("%d", count);
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 2, 4, 6, 4, 2 };
 
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    countSubarrays(arr, n);
    return 0;
}
 
// This code is contributed by piyush3010

Java




// Java program for the above approach
import java.util.*;
class GFG {
 
    // Function to count subarrays in
    // which sum of elements at even
    // and odd positions are equal
    static void countSubarrays(int arr[],
                               int n)
    {
        // Initialize variables
        int count = 0;
 
        // Iterate over the array
        for (int i = 0; i < n; i++) {
            int sum = 0;
 
            for (int j = i; j < n; j++) {
 
                // Check if position is
                // even then add to sum
                // then add it to sum
                if ((j - i) % 2 == 0)
                    sum += arr[j];
 
                // else subtract it to sum
                else
                    sum -= arr[j];
 
                // Increment the count
                // if the sum equals 0
                if (sum == 0)
 
                    count++;
            }
        }
 
        // Print the count of subarrays
        System.out.println(count);
    }
 
    // Driver Code
    public static void
        main(String[] args)
    {
        // Given array arr[]
        int arr[] = { 2, 4, 6, 4, 2 };
 
        // Size of the array
        int n = arr.length;
 
        // Function call
        countSubarrays(arr, n);
    }
}

Python3




# Python3 program for the above approach
 
# Function to count subarrays in
# which sum of elements at even
# and odd positions are equal
def countSubarrays(arr, n):
 
    # Initialize variables
    count = 0
 
    # Iterate over the array
    for i in range(n):
        sum = 0
         
        for j in range(i, n):
 
            # Check if position is
            # even then add to sum
            # hen add it to sum
            if ((j - i) % 2 == 0):
                sum += arr[j]
 
            # else subtract it to sum
            else:
                sum -= arr[j]
 
            # Increment the count
            # if the sum equals 0
            if (sum == 0):
                count += 1
                 
    # Print the count of subarrays
    print(count)
 
# Driver Code
if __name__ == '__main__':
   
    # Given array arr[]
    arr = [ 2, 4, 6, 4, 2 ]
 
    # Size of the array
    n = len(arr)
 
    # Function call
    countSubarrays(arr, n)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
static void countSubarrays(int []arr, int n)
{
     
    // Initialize variables
    int count = 0;
 
    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        int sum = 0;
 
        for(int j = i; j < n; j++)
        {
             
            // Check if position is
            // even then add to sum
            // then add it to sum
            if ((j - i) % 2 == 0)
                sum += arr[j];
 
            // else subtract it to sum
            else
                sum -= arr[j];
 
            // Increment the count
            // if the sum equals 0
            if (sum == 0)
                count++;
        }
    }
 
    // Print the count of subarrays
    Console.WriteLine(count);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 2, 4, 6, 4, 2 };
 
    // Size of the array
    int n = arr.Length;
 
    // Function call
    countSubarrays(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// javascript program for the above approach
 
// Function to count subarrays in
// which sum of elements at even
// and odd positions are equal
function countSubarrays(arr, n)
{
     
    // Initialize variables
    var count = 0;
    var i,j;
    // Iterate over the array
    for(i = 0; i < n; i++)
    {
        var sum = 0;
 
        for(j = i; j < n; j++)
        {
             
            // Check if position is
            // even then add to sum
            // then add it to sum
            if ((j - i) % 2 == 0)
                sum += arr[j];
 
            // Else subtract it to sum
            else
                sum -= arr[j];
 
            // Increment the count
            // if the sum equals 0
            if (sum == 0)
                count++;
        }
    }
 
    // Print the count of subarrays
    document.write(count);
}
 
// Driver Code
     
    // Given array arr[]
    var arr = [2, 4, 6, 4, 2];
 
    // Size of the array
    var n = arr.length;
 
    // Function call
    countSubarrays(arr, n);
 
</script>
Output: 
2

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :