Given an array of n positive numbers, the task is to count number of ordered pairs with even and odd sum.
Examples:
Input: arr[] = {1, 2, 4}
Output: Even sum Pairs = 2, Odd sum Pairs = 4
The ordered pairs are (1, 2), (1, 4), (2, 1), (4, 1), (2, 4), (4, 2)
Pairs with Even sum: (2, 4), (4, 2)
Pairs with Odd sum: (1, 2), (1, 4), (2, 1), (4, 1)Input: arr[] = {2, 4, 5, 9, 1, 8}
Output: Even sum Pairs = 12, Odd sum Pairs = 18
Approach:
The sum of two numbers is odd if one number is odd and another one is even. So now we have to count the even and odd numbers. As in order pair (a, b) and (b, a) both treated as different pair, therefore
Number of odd sum pairs = (count of even numbers) * (count of odd numbers) * 2
This is because every even number can pair with every odd number, and every odd number can pair with every even number. Thus multiply 2 is done in the answer.
And the number of even sum pairs will be an inversion of the number of odd sum pairs. Therefore:
Number of even sum pairs = Total Number of pairs – Number of odd sum pairs
Below is the implementation of above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// function to count odd sum pairint count_odd_pair(int n, int a[]){ int odd = 0, even = 0; for (int i = 0; i < n; i++) { // if number is even if (a[i] % 2 == 0) even++; // if number is odd else odd++; } // count of ordered pairs int ans = odd * even * 2; return ans;}// function to count even sum pairint count_even_pair(int odd_sum_pairs, int n){ int total_pairs = (n * (n - 1)); int ans = total_pairs - odd_sum_pairs; return ans;}// Driver codeint main(){ int n = 6; int a[] = { 2, 4, 5, 9, 1, 8 }; int odd_sum_pairs = count_odd_pair(n, a); int even_sum_pairs = count_even_pair( odd_sum_pairs, n); cout << "Even Sum Pairs = " << even_sum_pairs << endl; cout << "Odd Sum Pairs= " << odd_sum_pairs << endl; return 0;} |
Java
// Java implementation of the above approachclass GFG{ // function to count odd sum pair static int count_odd_pair(int n, int a[]) { int odd = 0, even = 0; for (int i = 0; i < n; i++) { // if number is even if (a[i] % 2 == 0) even++; // if number is odd else odd++; } // count of ordered pairs int ans = odd * even * 2; return ans; } // function to count even sum pair static int count_even_pair(int odd_sum_pairs, int n) { int total_pairs = (n * (n - 1)); int ans = total_pairs - odd_sum_pairs; return ans; } // Driver code public static void main(String []args) { int n = 6; int []a = { 2, 4, 5, 9, 1, 8 }; int odd_sum_pairs = count_odd_pair(n, a); int even_sum_pairs = count_even_pair( odd_sum_pairs, n); System.out.println("Even Sum Pairs = " + even_sum_pairs); System.out.println("Odd Sum Pairs= " + odd_sum_pairs); }}// This code is contributed by ihritik |
Python3
# Python3 implementation of the above approach# function to count odd sum pairdef count_odd_pair( n, a): odd = 0 even = 0 for i in range(0,n): # if number is even if ( a[ i] % 2 == 0): even=even+1 # if number is odd else: odd=odd+1 # count of ordered pairs ans = odd * even * 2 return ans # function to count even sum pairdef count_even_pair( odd_sum_pairs, n): total_pairs = ( n * ( n - 1)) ans = total_pairs - odd_sum_pairs return ans # Driver coden = 6a = [2, 4, 5, 9, 1, 8] odd_sum_pairs = count_odd_pair( n, a) even_sum_pairs = count_even_pair( odd_sum_pairs, n) print("Even Sum Pairs =", even_sum_pairs)print("Odd Sum Pairs=", odd_sum_pairs) # This code is contributed by ihritik |
C#
// C# implementation of the above approachusing System;class GFG{ // function to count odd sum pair static int count_odd_pair(int n, int []a) { int odd = 0, even = 0; for (int i = 0; i < n; i++) { // if number is even if (a[i] % 2 == 0) even++; // if number is odd else odd++; } // count of ordered pairs int ans = odd * even * 2; return ans; } // function to count even sum pair static int count_even_pair(int odd_sum_pairs, int n) { int total_pairs = (n * (n - 1)); int ans = total_pairs - odd_sum_pairs; return ans; } // Driver code public static void Main() { int n = 6; int []a = { 2, 4, 5, 9, 1, 8 }; int odd_sum_pairs = count_odd_pair(n, a); int even_sum_pairs = count_even_pair( odd_sum_pairs, n); Console.WriteLine("Even Sum Pairs = " + even_sum_pairs); Console.WriteLine("Odd Sum Pairs= " + odd_sum_pairs); }}// This code is contributed by ihritik |
PHP
<?php// PHP implementation of the above approach// function to count odd sum pairfunction count_odd_pair($n, $a){ $odd = 0; $even = 0; for ($i = 0; $i < $n; $i++) { // if number is even if ($a[$i] % 2 == 0) $even++; // if number is odd else $odd++; } // count of ordered pairs $ans = $odd * $even * 2; return $ans;}// function to count even sum pairfunction count_even_pair($odd_sum_pairs, $n){ $total_pairs = ($n * ($n - 1)); $ans = $total_pairs - $odd_sum_pairs; return $ans;}// Driver code$n = 6;$a = array( 2, 4, 5, 9, 1, 8 );$odd_sum_pairs = count_odd_pair($n, $a);$even_sum_pairs = count_even_pair($odd_sum_pairs, $n);echo "Even Sum Pairs = $even_sum_pairs \n";echo "Odd Sum Pairs= $odd_sum_pairs \n"; // This code is contributed by ihritik ?> |
Javascript
<script>// JavaScript implementation of the above approach// function to count odd sum pairfunction count_odd_pair(n, a){ var odd = 0, even = 0; for (var i = 0; i < n; i++) { // if number is even if (a[i] % 2 == 0) even++; // if number is odd else odd++; } // count of ordered pairs var ans = odd * even * 2; return ans;}// function to count even sum pairfunction count_even_pair(odd_sum_pairs, n){ var total_pairs = (n * (n - 1)); var ans = total_pairs - odd_sum_pairs; return ans;}// Driver codevar n = 6;var a = [2, 4, 5, 9, 1, 8];var odd_sum_pairs = count_odd_pair(n, a);var even_sum_pairs = count_even_pair( odd_sum_pairs, n); document.write( "Even Sum Pairs = " + even_sum_pairs + "<br>"); document.write( "Odd Sum Pairs = " + odd_sum_pairs + "<br>"); </script> |
Output
Even Sum Pairs = 12 Odd Sum Pairs= 18
Complexity Analysis:
- Time Complexity: O(n), to find the number of odd and even numbers in the given array
- Auxiliary Space: O(1), as no extra space is used
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
