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# Count number of ordered pairs with Even and Odd Sums

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 ``using` `namespace` `std;` `// function to count odd sum pair``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``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``int` `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 approach` `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(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 pair``def` `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 pair``def` `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 ``=` `[``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 approach` `using` `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

 ``

## Javascript

 ``

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

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