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

Last Updated : 11 Jul, 2022

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:

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

Count of subarrays of size K with elements having even frequencies

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

C++

C

Java

Python3

C#

Javascript

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