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)
Related Topic: Subarrays, Subsequences, and Subsets in Array