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Count pairs in an array containing at least one even value

  • Last Updated : 26 Jul, 2021

Given an array arr[], the task is to count pairs such that each pair (arr[i], arr[j]) contains at least one even element in it where i != j.

Examples: 

Input: arr[] = {1, 2, 3, 1, 3} 
Output:
Explanation: 
Possible pairs are: (1, 2), (2, 3), (2, 1), (2, 3).

Input: arr[] = {8, 2, 3, 1, 4, 2} 
Output: 14 
Explanation: 
Possible pairs are: (8, 2), (8, 3), (8, 1), (8, 4), (8, 2), (2, 3), (2, 1), (2, 4), (2, 2), (3, 4), (3, 2), (1, 4), (1, 2), (4, 2). 

A Simple Approach is to run two loops. Pick each element one-by-one and for each element find element on right side of array that holds condition, then increment count.
Time Complexity: 



O(N^{2})

Below is the implementation of the above approach:   

C++




// C++ implementation to count
// pairs in an array such that
// each pair contains at
// least one even element
#include<bits/stdc++.h>
using namespace std;
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
int CountPairs(int arr[], int n)
{
    int count = 0;
  
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(int i = 0; i < n; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 8, 2, 3, 1, 4, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    // Function call
    cout << (CountPairs(arr, n));
}
  
// This code is contributed by rock_cool

Java




// Java implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
import java.util.*;
 
class GFG {
 
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    static int CountPairs(int[] arr, int n)
    {
 
        int count = 0;
 
        // Generate all possible pairs
        // and increment then count
        // if the condition is satisfied
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
 
                if (arr[i] % 2 == 0
                    || arr[j] % 2 == 0)
                    count++;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int[] arr = { 8, 2, 3, 1, 4, 2 };
        int n = arr.length;
 
        // Function Call
        System.out.println(CountPairs(arr, n));
    }
}

Python3




# Python3 implementation to count
# pairs in an array such that
# each pair contains at
# least one even element
def CountPairs(arr, n):
     
    count = 0
     
    # Generate all possible pairs
    # and increment then count
    # if the condition is satisfied
    for i in range(n):
        for j in range(i + 1, n):
            if (arr[i] % 2 == 0 or
                arr[j] % 2 == 0):
                count += 1
                 
    return count
     
# Driver code
arr = [ 8, 2, 3, 1, 4, 2 ]
n = len(arr)
 
# Function call
print(CountPairs(arr, n))
 
# This code is contributed by rutvik_56

C#




// C# implementation to count
// pairs in an array such that
// each pair contains at
// least one even element
using System;
  
class GFG{
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
static int CountPairs(int[] arr, int n)
{
  
    int count = 0;
  
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(int i = 0; i < n; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
  
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 8, 2, 3, 1, 4, 2 };
    int n = arr.Length;
  
    // Function Call
    Console.WriteLine(CountPairs(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript implementation to count
// pairs in an array such that each
// pair contains at least one even element
 
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
function CountPairs(arr, n)
{
    let count = 0;
 
    // Generate all possible pairs
    // and increment then count
    // if the condition is satisfied
    for(let i = 0; i < n; i++)
    {
       for(let j = i + 1; j < n; j++)
       {
          if (arr[i] % 2 == 0 ||
              arr[j] % 2 == 0)
              count++;
       }
    }
    return count;
}
 
// Driver code
let arr = [ 8, 2, 3, 1, 4, 2 ];
let n = arr.length;
 
// Function call
document.write(CountPairs(arr, n));
 
// This code is contributed by divyeshrabadiya07
 
</script>
Output: 
14

 

Efficient Approach: The idea is to count the even and odd elements in the array and include pairs having only one even element or both the pairs to be even element. 

  • Pair having exactly one even element: count of the pairs having exactly one even element will be:

(even*odd)

  • Pair having exactly two even elements: count of the pairs having exactly two even elements will be:

\frac{(even*(even-1))}{2}        

Therefore, the count of the pairs having at least one even element will be 
\frac{(even*(even-1))}{2} + (even*odd)

Below is the implementation of the above approach: 

C++




// C++ implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
int CountPairs(int arr[], int n)
{
     
    // Store count of even
    // and odd elements
    int even = 0, odd = 0;
    for(int i = 0; i < n; i++)
    {
         
       // Check element is
       // even or odd
       if (arr[i] % 2 == 0)
           even++;
       else
           odd++;
    }
    return (even * (even - 1)) / 2 +
           (even * odd);
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 2, 3, 1, 4, 2 };
    int n = sizeof(arr) / sizeof(int);
     
    cout << CountPairs(arr, n);
}
 
// This code is contributed by jrishabh99

Java




// Java implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
import java.util.*;
 
class GFG {
 
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    static int CountPairs(int[] arr, int n)
    {
        // store count of even
        // and odd elements
        int even = 0, odd = 0;
 
        for (int i = 0; i < n; i++) {
 
            // check element is
            // even or odd
            if (arr[i] % 2 == 0)
                even++;
            else
                odd++;
        }
 
        return (even * (even - 1)) / 2
            + (even * odd);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[] arr = { 8, 2, 3, 1, 4, 2 };
        int n = arr.length;
        System.out.println(CountPairs(arr, n));
    }
}

Python3




# Python3 implementation to count
# pairs in an array such that
# each pair contains at
# least one even element
   
# Function to count the pairs in
# the array such as there is at
# least one even element in each pair
def CountPairs(arr, n):
       
    # Store count of even
    # and odd elements
    even = 0
    odd = 0
     
    for i in range(n):
           
       # Check element is
       # even or odd
       if (arr[i] % 2 == 0):
           even += 1
       else:
           odd += 1
      
    return ((even * (even - 1)) // 2 +
            (even * odd))
  
# Driver Code
arr = [ 8, 2, 3, 1, 4, 2 ]
n = len(arr)
       
print(CountPairs(arr, n))
 
# This code is contributed by code_hunt

C#




// C# implementation to Count
// pairs in an array such that
// each pair contains at
// least one even element
using System;
class GFG{
  
// Function to count the pairs in
// the array such as there is at
// least one even element in each pair
static int CountPairs(int[] arr, int n)
{
      
    // Store count of even
    // and odd elements
    int even = 0, odd = 0;
  
    for(int i = 0; i < n; i++)
    {
         
       // Check element is
       // even or odd
       if (arr[i] % 2 == 0)
           even++;
       else
           odd++;
    }
    return (even * (even - 1)) / 2 +
           (even * odd);
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 8, 2, 3, 1, 4, 2 };
    int n = arr.Length;
      
    Console.Write(CountPairs(arr, n));
}
}
  
// This code is contributed by Nidhi_biet

Javascript




<script>
    // Javascript implementation to Count
    // pairs in an array such that
    // each pair contains at
    // least one even element
     
    // Function to count the pairs in
    // the array such as there is at
    // least one even element in each pair
    function CountPairs(arr, n)
    {
 
        // Store count of even
        // and odd elements
        let even = 0, odd = 0;
        for(let i = 0; i < n; i++)
        {
 
           // Check element is
           // even or odd
           if (arr[i] % 2 == 0)
               even++;
           else
               odd++;
        }
        return (even * (even - 1)) / 2 + (even * odd);
    }
     
    let arr = [ 8, 2, 3, 1, 4, 2 ];
    let n = arr.length;
      
    document.write(CountPairs(arr, n));
     
    // This code is contributed by divyesh072019.
</script>
Output: 
14

 

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

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