Given an array **arr[]** of size **N**, the task is to replace a pair of array elements whose Bitwise XOR is even by their** Bitwise XOR**. Repeat the above step as long as possible. Finally, print the count of such operations performed on the array

**Examples:**

Input:arr[] = { 4, 6, 1, 3 }Output:3Explanation:

Step 1: Remove the pair (4, 6) and replace them by their XOR value (= 2) in the array. Therefore, the array arr[] modifies to {2, 1, 3}.

Step 2: Remove the pair (1, 3) and replace them by their XOR value (= 2) in the array, modifies the array as arr[] = {2, 2}.

At last select the pair (2, 2) and then remove the pair and insert the xor of 2 and 2 in the array which modifies the array as arr[] ={0}.

Now no other pair can be chosen therefore 3 is the maximum number of pairs whose Xor is even.

Input:arr[ ] = { 1, 2, 3, 4, 5 }Output:3

**Naive Approach:** The simplest approach to solve this problem is to find all possible pairs of the array and for each pair, check if their Bitwise XOR is even or not. If found to be true, then increment the count of pairs and remove both the elements from the array and add their XOR to the array. Repeat the above steps until no more pairs can be selected. Print the count of operations performed. **Time Complexity:** O(N^{3})**Auxiliary Space:** O(1)

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

Even ^ Even = EvenOdd ^ Odd = EvenThe total number of pairs that can be formed from only odd numbers satisfying the conditions isodd / 2.The total numbers of pairs that can be formed from only even numbers satisfying the conditions iseven – 1.

Follow the steps below to solve the problem:

- Traverse the array.
- Count the frequency of odd numbers and store it in a variable, say
**odd**. - The total number of pairs with even XOR that can be formed from all the odd array elements is
**floor(odd / 2)**. - Deleting the formed pairs in the above step and replacing them with their XOR values respectively, increases the count of even elements by
**floor(odd / 2).** - Finally, print the count of pairs that can be formed with even XOR as
**(N – odd + odd/2 -1) + odd / 2.**

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 maximize the count` `// of pairs with even XOR possible` `// in an array by given operations` `int` `countPairs(` `int` `arr[], ` `int` `N)` `{` ` ` `// Stores count of odd` ` ` `// array elements` ` ` `int` `odd = 0;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If arr[i] is odd` ` ` `if` `(arr[i] & 1)` ` ` `odd++;` ` ` `}` ` ` `// Stores the total number` ` ` `// of even pairs` ` ` `int` `ans = (N - odd + odd / 2` ` ` `- 1)` ` ` `+ odd / 2;` ` ` `return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Input` ` ` `int` `arr[] = { 4, 6, 1, 3 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function call to count the number` ` ` `// of pairs whose XOR is even` ` ` `cout << countPairs(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement the above approach` `public` `class` `GFG` `{` ` ` `// Function to maximize the count` ` ` `// of pairs with even XOR possible` ` ` `// in an array by given operations` ` ` `static` `int` `countPairs(` `int` `[]arr, ` `int` `N)` ` ` `{` ` ` `// Stores count of odd` ` ` `// array elements` ` ` `int` `odd = ` `0` `;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) ` ` ` `{` ` ` `// If arr[i] is odd` ` ` `if` `((arr[i] & ` `1` `)!=` `0` `)` ` ` `odd++;` ` ` `}` ` ` `// Stores the total number` ` ` `// of even pairs` ` ` `int` `ans = (N - odd + odd / ` `2` ` ` `- ` `1` `)` ` ` `+ odd / ` `2` `;` ` ` `return` `ans;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `// Input` ` ` `int` `[]arr = { ` `4` `, ` `6` `, ` `1` `, ` `3` `};` ` ` `int` `N = arr.length;` ` ` `// Function call to count the number` ` ` `// of pairs whose XOR is even` ` ` `System.out.println(countPairs(arr, N));` ` ` `}` `}` `// This code is contributed by AnkThon.` |

## Python3

`# Python3 program to implement the above approach` `# Function to maximize the count` `# of pairs with even XOR possible` `# in an array by given operations` `def` `countPairs(arr, N):` ` ` ` ` `# Stores count of odd` ` ` `# array elements` ` ` `odd ` `=` `0` ` ` `# Traverse the array` ` ` `for` `i ` `in` `range` `(N):` ` ` `# If arr[i] is odd` ` ` `if` `(arr[i] & ` `1` `):` ` ` `odd ` `+` `=` `1` ` ` `# Stores the total number` ` ` `# of even pairs` ` ` `ans ` `=` `(N ` `-` `odd ` `+` `odd ` `/` `/` `2` `-` `1` `) ` `+` `odd ` `/` `/` `2` ` ` `return` `ans` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Input` ` ` `arr ` `=` `[` `4` `, ` `6` `, ` `1` `, ` `3` `]` ` ` `N ` `=` `len` `(arr)` ` ` `# Function call to count the number` ` ` `# of pairs whose XOR is even` ` ` `print` `(countPairs(arr, N))` ` ` `# This code is contributed by mohit kumar 29.` |

## C#

`// C# program to implement the above approach` `using` `System;` `class` `GFG` `{` `// Function to maximize the count` `// of pairs with even XOR possible` `// in an array by given operations` `static` `int` `countPairs(` `int` `[]arr, ` `int` `N)` `{` ` ` ` ` `// Stores count of odd` ` ` `// array elements` ` ` `int` `odd = 0;` ` ` `// Traverse the array` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `{` ` ` `// If arr[i] is odd` ` ` `if` `((arr[i] & 1)!=0)` ` ` `odd++;` ` ` `}` ` ` `// Stores the total number` ` ` `// of even pairs` ` ` `int` `ans = (N - odd + odd / 2` ` ` `- 1)` ` ` `+ odd / 2;` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` ` ` `// Input` ` ` `int` `[]arr = { 4, 6, 1, 3 };` ` ` `int` `N = arr.Length;` ` ` `// Function call to count the number` ` ` `// of pairs whose XOR is even` ` ` `Console.Write(countPairs(arr, N));` `}` `}` `// This code is contributed by SURENDRA_GANGWAR.` |

**Output:**

3

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

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