Count even sum pairs possible by selecting two integers from two given ranges respectively

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Given two positive integers X and Y, the task is to count pairs with even sum possible by choosing any integer from the range 1 to X and another integer from the range 1 to Y.

Examples:

Input : X = 2, Y = 3
Output : 3
Explanation : All such possible pairs are {1, 1}, {1, 3}, {2, 2}.

Approach: Follow the steps below to solve the problem:

Initialize a variable, say cntXEvenNums, to store the count of even numbers in the range [1, X] obtained by dividing X by 2.
Initialize a variable, say cntXOddNums, to store the count of odd numbers in the range [1, X] obtained by dividing (X + 1) by 2.
Initialize a variable, say cntYEvenNums, to store the count of even numbers between 1 to Y by dividing Y by 2.
Initialize a variable, say cntYOddNums, to store the count of odd numbers between 1 to Y by dividing (Y + 1) by 2.
Initialize a variable, say cntPairs, to store the count of even sum pairs by multiplying cntXEvenNums with cntYEvenNums and multiplying cntXOddNums with cntYOddNums and find the sum of both.
Finally, print the value of cntPairs obtained.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count even
// sum pairs in the given range
long long cntEvenSumPairs(long long X, long long Y)
{
    // Stores the count of even
    // numbers between 1 to X
    long long cntXEvenNums = X / 2;
 
    // Stores the count of odd
    // numbers between 1 to X
    long long cntXOddNums = (X + 1) / 2;
 
    // Stores the count of even
    // numbers between 1 to Y
    long long cntYEvenNums = Y / 2;
 
    // Stores the count of odd
    // numbers between 1 to Y
    long long cntYOddNums = (Y + 1) / 2;
 
    // Stores the count of
    // pairs having even sum
    long long cntPairs
        = (cntXEvenNums * 1LL * cntYEvenNums)
          + (cntXOddNums * 1LL * cntYOddNums);
 
    // Returns the count of pairs
    // having even sum
    return cntPairs;
}
 
// Driver Code
int main()
{
    long long X = 2;
    long long Y = 3;
 
    cout << cntEvenSumPairs(X, Y);
    return 0;
}

Java

// Java program to implement
// the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to count maximum  even
    // sum pairs in the given range
    static long cntEvenSumPairs(long X, long Y)
    {
 
        // Stores the count of even
        // numbers between 1 to X
        long cntXEvenNums = X / 2;
 
        // Stores the count of odd
        // numbers between 1 to X
        long cntXOddNums = (X + 1) / 2;
 
        // Stores the count of even
        // numbers between 1 to Y
        long cntYEvenNums = Y / 2;
 
        // Stores the count of odd
        // numbers between 1 to Y
        long cntYOddNums = (Y + 1) / 2;
 
        // Stores the count of
        // pairs having even sum
        long cntPairs = (cntXEvenNums * cntYEvenNums)
                        + (cntXOddNums * cntYOddNums);
 
        // Returns the count of pairs
        // having even sum
        return cntPairs;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        long X = 2;
        long Y = 3;
 
        System.out.println(cntEvenSumPairs(X, Y));
    }
}

Python

# Python program to implement 
# the above approach 
 
# Function to count even 
# sum pairs in the given range 
def cntEvenSumPairs(X, Y):
     
    # Stores the count of even
    # numbers between 1 to X
    cntXEvenNums = X / 2
 
    # Stores the count of odd
    # numbers between 1 to X
    cntXOddNums = (X + 1) / 2
 
    # Stores the count of even
    # numbers between 1 to Y
    cntYEvenNums = Y / 2
 
    # Stores the count of odd
    # numbers between 1 to Y
    cntYOddNums = (Y + 1) / 2
 
    # Stores the count of
    # pairs having even sum
    cntPairs = ((cntXEvenNums * cntYEvenNums) +
                 (cntXOddNums * cntYOddNums))
  
    # Returns the count of pairs
    # having even sum
    return cntPairs
 
# Driver code
X = 2
Y = 3
 
print(cntEvenSumPairs(X, Y))
 
# This code is contributed by hemanth gadarla

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count maximum  even
// sum pairs in the given range
static long cntEvenSumPairs(long X, long Y)
{
     
    // Stores the count of even
    // numbers between 1 to X
    long cntXEvenNums = X / 2;
 
    // Stores the count of odd
    // numbers between 1 to X
    long cntXOddNums = (X + 1) / 2;
 
    // Stores the count of even
    // numbers between 1 to Y
    long cntYEvenNums = Y / 2;
 
    // Stores the count of odd
    // numbers between 1 to Y
    long cntYOddNums = (Y + 1) / 2;
 
    // Stores the count of
    // pairs having even sum
    long cntPairs = (cntXEvenNums * cntYEvenNums) + 
                     (cntXOddNums * cntYOddNums);
 
    // Returns the count of pairs
    // having even sum
    return cntPairs;
}
 
// Driver Code
public static void Main(string[] args)
{
    long X = 2;
    long Y = 3;
 
    Console.WriteLine(cntEvenSumPairs(X, Y));
}
}
 
// This code is contributed by chitranayal

Javascript

<script>
 
// Javascript program to implement
// the above approach    
 
     // Function to count maximum even
    // sum pairs in the given range
    function cntEvenSumPairs(X , Y) {
 
        // Stores the count of even
        // numbers between 1 to X
        var cntXEvenNums = parseInt(X / 2);
 
        // Stores the count of odd
        // numbers between 1 to X
        var cntXOddNums = parseInt((X + 1) / 2);
 
        // Stores the count of even
        // numbers between 1 to Y
        var cntYEvenNums = parseInt(Y / 2);
 
        // Stores the count of odd
        // numbers between 1 to Y
        var cntYOddNums =parseInt( (Y + 1) / 2);
 
        // Stores the count of
        // pairs having even sum
        var cntPairs = (cntXEvenNums * cntYEvenNums) +
        (cntXOddNums * cntYOddNums);
 
        // Returns the count of pairs
        // having even sum
        return cntPairs;
    }
 
    // Driver Code
     
        var X = 2;
        var Y = 3;
 
        document.write(cntEvenSumPairs(X, Y));
 
// This code contributed by Rajput-Ji 
 
</script>

Output:

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

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