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Count pair of integers having even sum
  • Last Updated : 13 Apr, 2021

Given two integers N and M, the task is to count all possible pair of integers (i, j) (1 ≤ i ≤ N, 1 ≤ j ≤ M) such that i + j is even.

Examples:

Input: N = 6, M = 4 
Output: 12 
Explanation: The pairs (1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2), (6, 4) satisfy the required condition. Therefore, the count is 12.

Input: N = 2 and M = 8 
Output:
 

Naive Approach: The simplest approach to solve this problem is to traverse over the range [1, M] for every number in that range, traverse the range [1, N] and generate all possible pairs. For every possible pair, check its sum is an even number or not. If found to be true, increment the count. Finally, print the count obtained.



Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to count pairs with even sum
int countEvenPairs(int N, int M)
{
     
    // Stores the count of pairs
    // with even sum
    int count = 0;
 
    // Traverse the range 1 to N
    for(int i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(int j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
    int M = 6;
     
    cout << countEvenPairs(N, M) << endl;
     
    return 0;
}
 
// This code is contributed by akhilsaini

Java




// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to count pairs with even sum
    public static int countEvenPairs(
        int N, int M)
    {
 
        // Stores the count of pairs
        // with even sum
        int count = 0;
 
        // Traverse the range 1 to N
        for (int i = 1; i <= N; i++) {
 
            // Traverse the range 1 to M
            for (int j = 1; j <= M; j++) {
 
                // Check if the sum of the
                // pair (i, j) is even or not
                if ((i + j) % 2 == 0) {
 
                    // Update count
                    count++;
                }
            }
        }
 
        // Return the count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        int M = 6;
        System.out.print(
            countEvenPairs(N, M));
    }
}

Python3




# Python3 program for the above approach
 
# Function to count pairs with even sum
def countEvenPairs(N, M):
     
    # Stores the count of pairs
    # with even sum
    count = 0
     
    # Traverse the range 1 to N
    for i in range(1, N + 1):
     
        # Traverse the range 1 to M
        for j in range(1, M + 1):
         
            # Check if the sum of the
            # pair (i, j) is even or not
            if ((i + j) % 2 == 0):
               
                # Update count
                count += 1
     
    # Return the count
    return count
 
# Driver Code
if __name__ == '__main__':
     
  N = 4
  M = 6
   
  print(countEvenPairs(N, M))
   
# This code is contributed by akhilsaini

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to count pairs with even sum
public static int countEvenPairs(int N, int M)
{
     
    // Stores the count of pairs
    // with even sum
    int count = 0;
 
    // Traverse the range 1 to N
    for(int i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(int j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
     
    // Return the count
    return count;
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    int M = 6;
     
    Console.WriteLine(
        countEvenPairs(N, M));
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to count pairs with even sum
function countEvenPairs(N, M)
{
     
    // Stores the count of pairs
    // with even sum
    var count = 0;
 
    // Traverse the range 1 to N
    for(var i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(var j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver code
var N = 4;
var M = 6;
 
document.write(countEvenPairs(N, M));
 
// This code is contributed by Kirti
 
</script>
Output: 
12

 

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

Efficient Approach: The above approach can be optimized based on the following observations: 

  • Even number + Even number = Even number
  • Odd number + Odd number = Even number

Follow the steps below to solve the problem: 

  • Initialize two variables, say nEven and nOdd, to store the count of odd and even integers up to N.
  • Initialize two variables, say mEven and mOdd, to store the count of even and odd integers up to M.
  • Finally, count the required number of pairs uing the formula:

count = nEven * mEven + nOdd * mOdd

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 even pairs
int countEvenPairs(int N, int M)
{
 
    // Stores count of pairs having even sum
    int count = 0;
 
    // Stores count of even numbers up to N
    int nEven = floor(N / 2);
 
    // Stores count of odd numbers up to N
    int nOdd = ceil(N / 2);
 
    // Stores count of even numbers up to M
    int mEven = floor(M / 2);
 
    // Stores count of odd numbers up to M
    int mOdd = ceil(M / 2);
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
    int M = 6;
    cout << countEvenPairs(N, M);
    return 0;
}
 
// This code is contributed by Dharanendra L V

Java




// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to count even pairs
    public static int countEvenPairs(
        int N, int M)
    {
 
        // Stores count of pairs having even sum
        int count = 0;
 
        // Stores count of even numbers up to N
        int nEven = (int)Math.floor((double)N / 2);
 
        // Stores count of odd numbers up to N
        int nOdd = (int)Math.ceil((double)N / 2);
 
        // Stores count of even numbers up to M
        int mEven = (int)Math.floor((double)M / 2);
 
        // Stores count of odd numbers up to M
        int mOdd = (int)Math.ceil((double)M / 2);
 
        count = nEven * mEven + nOdd * mOdd;
 
        // Return the count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        int M = 6;
        System.out.print(countEvenPairs(N, M));
    }
}

Python3




# Python3 program for the above approach
import math
 
# Function to count even pairs
def countEvenPairs(N, M):
   
    # Stores count of pairs having even sum
    count = 0;
 
    # Stores count of even numbers up to N
    nEven = int(math.floor(N / 2));
 
    # Stores count of odd numbers up to N
    nOdd = int(math.ceil(N / 2));
 
    # Stores count of even numbers up to M
    mEven = int(math.floor(M / 2));
 
    # Stores count of odd numbers up to M
    mOdd = int(math.ceil(M / 2));
 
    count = nEven * mEven + nOdd * mOdd;
 
    # Return the count
    return count;
 
# Driver Code
if __name__ == '__main__':
    N = 4;
    M = 6;
    print(countEvenPairs(N, M));
 
# This code is contributed by 29AjayKumar

C#




// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to count even pairs
  public static int countEvenPairs(int N, int M)
  {
 
    // Stores count of pairs having even sum
    int count = 0;
 
    // Stores count of even numbers up to N
    int nEven = (int)Math.Floor((double)N / 2);
 
    // Stores count of odd numbers up to N
    int nOdd = (int)Math.Ceiling((double)N / 2);
 
    // Stores count of even numbers up to M
    int mEven = (int)Math.Floor((double)M / 2);
 
    // Stores count of odd numbers up to M
    int mOdd = (int)Math.Ceiling((double)M / 2);
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 4;
    int M = 6;
    Console.Write(countEvenPairs(N, M));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to count even pairs
function countEvenPairs(N, M)
{
     
    // Stores count of pairs having even sum
    let count = 0;
 
    // Stores count of even numbers up to N
    nEven = parseInt(Math.floor(N / 2));
 
    // Stores count of odd numbers up to N
    nOdd = parseInt(Math.ceil(N / 2));
 
    // Stores count of even numbers up to M
    mEven = parseInt(Math.floor(M / 2));
 
    // Stores count of odd numbers up to M
    mOdd = parseInt(Math.ceil(M / 2));
 
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
}
 
// Driver Code
let N = 4;
let M = 6;
 
document.write(countEvenPairs(N, M));
 
// This code is contributed by mohan1240760
 
</script>
Output: 
12

 

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

 

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