# Count of all possible pairs of array elements with same parity

Given an array A[] of integers, the task is to find the total number of pairs such that each pair contains either both even or both odd elements. A vaild pair (A[ i ], A[ j ]) can only be formed if i != j.

Examples:

Input: A[ ] = {1, 2, 3, 1, 3}
Output: 6
Explanation:
Possible odd pairs = (1, 3), (1, 1), (1, 3), (3, 1), (3, 3), (1, 3) = 6
Possible even pairs = 0
Hence, total pairs = 6 + 0 = 6

Input: A[ ] = {8, 2, 3, 1, 4, 2}
Output: 7
Explanation:
Possible odd pair = (3, 1) = 1
Possible even pairs = (8, 2), (8, 4), (8, 2), (2, 4), (2, 2), (4, 2) = 6
Hence, total pairs = 6 + 1 = 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach:
The simplest approach is to generate all possible pairs. For each pair, check if both elements are odd or both are even. If so, increment a counter. The final count will be the required answer.
Time Complexity: O(N2)

Below is the implementation of the above approach:

## Java

 `// Java program for above approach ` ` `  `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `countPairs( ` `        ``int``[] A, ``int` `n) ` `    ``{ ` `        ``int` `count = ``0``, i, j; ` ` `  `        ``// Generate all possible pairs ` `        ``for` `(i = ``0``; i < n; i++) { ` `            ``for` `(j = i + ``1``; j < n; j++) { ` ` `  `                ``// Increment the count if ` `                ``// both even or both odd ` `                ``if` `((A[i] % ``2` `== ``0` `                     ``&& A[j] % ``2` `== ``0``) ` `                    ``|| (A[i] % ``2` `!= ``0` `                        ``&& A[j] % ``2` `!= ``0``)) ` `                    ``count++; ` `            ``} ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main( ` `        ``String[] args) ` `    ``{ ` `        ``int``[] A = { ``1``, ``2``, ``3``, ``1``, ``3` `}; ` `        ``int` `n = A.length; ` `        ``System.out.println( ` `            ``countPairs(A, n)); ` `    ``} ` `} `

## C#

 `// C# program for above approach ` `using` `System; ` `class` `GFG{ ` ` `  `static` `int` `countPairs(``int``[] A, ``int` `n) ` `{ ` `    ``int` `count = 0, i, j; ` ` `  `    ``// Generate all possible pairs ` `    ``for` `(i = 0; i < n; i++) ` `    ``{ ` `        ``for` `(j = i + 1; j < n; j++)  ` `        ``{ ` ` `  `            ``// Increment the count if ` `            ``// both even or both odd ` `            ``if` `((A[i] % 2 == 0 && A[j] % 2 == 0) ||  ` `                ``(A[i] % 2 != 0 && A[j] % 2 != 0)) ` `                ``count++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int``[] A = { 1, 2, 3, 1, 3 }; ` `    ``int` `n = A.Length; ` `    ``Console.Write(countPairs(A, n)); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

Output:

```6
```

Efficient Approach:
Traverse the array and count and store even and odd numbers in the array and calculate the possible pairs from respective counts and display their sum.

Let the count of even and odd elements in the array be EC and OC respectively.
Count of even pairs = ( EC * ( EC – 1 ) ) / 2
Count of odd pairs = ( OC * ( OC – 1 ) ) / 2
Hence, total number of possible pairs = (( EC * ( EC – 1 ) ) + ( OC * ( OC – 1 ) ))/ 2

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach ` ` `  `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``static` `int` `countPairs( ` `        ``int``[] A, ``int` `n) ` `    ``{ ` `        ``// Store count of ` `        ``// even and odd elements ` `        ``int` `even = ``0``, odd = ``0``; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``if` `(A[i] % ``2` `== ``0``) ` `                ``even++; ` `            ``else` `                ``odd++; ` `        ``} ` ` `  `        ``return` `(even * (even - ``1``)) / ``2` `            ``+ (odd * (odd - ``1``)) / ``2``; ` `    ``} ` ` `  `    ``// Driver Program ` `    ``public` `static` `void` `main( ` `        ``String[] args) ` `    ``{ ` ` `  `        ``int``[] A = { ``1``, ``2``, ``3``, ``1``, ``3` `}; ` `        ``int` `n = A.length; ` `        ``System.out.println( ` `            ``countPairs(A, n)); ` `    ``} ` `} `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `static` `int` `countPairs(``int``[] A, ``int` `n) ` `{ ` `     `  `    ``// Store count of ` `    ``// even and odd elements ` `    ``int` `even = 0, odd = 0; ` ` `  `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{ ` `       ``if` `(A[i] % 2 == 0) ` `           ``even++; ` `       ``else` `           ``odd++; ` `    ``} ` ` `  `    ``return` `(even * (even - 1)) / 2 +  ` `            ``(odd * (odd - 1)) / 2; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] A = { 1, 2, 3, 1, 3 }; ` `    ``int` `n = A.Length; ` `     `  `    ``Console.Write(countPairs(A, n)); ` `} ` `} ` ` `  `// This code is contributed by nidhi_biet `

Output:

```6
```

Time complexity: O(N)

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