# Count of even and odd set bit Array elements after XOR with K for Q queries

Given an array arr of N elements and another array Q containing values of K, the task is to print the count of elements in the array arr with odd and even set bits after its XOR with each element K in the array Q.

Examples:

Input: arr[] = { 2, 7, 4, 5, 3 }, Q[] = { 3, 4, 12, 6 }
Output: 2 3
3 2
2 3
2 3

Input: arr[] = { 7, 1, 6, 5, 11 }, Q[] = { 2, 10, 3, 6 }
Output: 3 2
2 3
2 3
2 3

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

Approach:

• XOR of two elements both having odd or even set bits, results to even set bits.
• XOR of two elements, one having odd and other having even set bits or vice versa, results to even set bits.
• Precompute count of elements with even and odd set bits of all array elements using Brian Kernighan’s Algorithm.
• For all elements of Q, count number of set bits. If count of set bits is even, the count of even and odd set bits elements remain unchanged. Otherwise reverse the count and display.

Below is the implementation of the above approach:

## C++

 `// C++ Program to count number ` `// of even and odd set bits ` `// elements after XOR with a ` `// given element ` ` `  `#include ` `using` `namespace` `std; ` ` `  `void` `keep_count(``int` `arr[], ``int``& even, ` `                ``int``& odd, ``int` `N) ` `{ ` `    ``// Store the count of set bits ` `    ``int` `count; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``count = 0; ` ` `  `        ``// Brian Kernighan's algorithm ` `        ``while` `(arr[i] != 0) { ` `            ``arr[i] = (arr[i] - 1) & arr[i]; ` `            ``count++; ` `        ``} ` ` `  `        ``if` `(count % 2 == 0) ` `            ``even++; ` `        ``else` `            ``odd++; ` `    ``} ` ` `  `    ``return``; ` `} ` ` `  `// Function to solve Q queries ` `void` `solveQueries( ` `    ``int` `arr[], ``int` `n, ` `    ``int` `q[], ``int` `m) ` `{ ` ` `  `    ``int` `even_count = 0, odd_count = 0; ` ` `  `    ``keep_count(arr, even_count, ` `               ``odd_count, n); ` ` `  `    ``for` `(``int` `i = 0; i < m; i++) { ` ` `  `        ``int` `X = q[i]; ` ` `  `        ``// Store set bits in X ` `        ``int` `count = 0; ` ` `  `        ``// Count set bits of X ` `        ``while` `(X != 0) { ` `            ``X = (X - 1) & X; ` `            ``count++; ` `        ``} ` ` `  `        ``if` `(count % 2 == 0) { ` `            ``cout << even_count << ``" "` `                 ``<< odd_count << ``"\n"``; ` `        ``} ` `        ``else` `{ ` `            ``cout << odd_count << ``" "` `                 ``<< even_count ` `                 ``<< ``"\n"``; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 7, 4, 5, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``int` `q[] = { 3, 4, 12, 6 }; ` `    ``int` `m = ``sizeof``(q) / ``sizeof``(q); ` ` `  `    ``solveQueries(arr, n, q, m); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to count number ` `// of even and odd set bits ` `// elements after XOR with a ` `// given element ` `class` `GFG{ ` `     `  `static` `int` `even, odd; ` ` `  `static` `void` `keep_count(``int` `arr[], ``int` `N) ` `{ ` `     `  `    ``// Store the count of set bits ` `    ``int` `count; ` `     `  `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `       ``count = ``0``; ` `        `  `       ``// Brian Kernighan's algorithm ` `       ``while` `(arr[i] != ``0``) ` `       ``{ ` `           ``arr[i] = (arr[i] - ``1``) & arr[i]; ` `           ``count++; ` `       ``} ` `       ``if` `(count % ``2` `== ``0``) ` `           ``even++; ` `       ``else` `           ``odd++; ` `    ``} ` `    ``return``; ` `} ` ` `  `// Function to solve Q queries ` `static` `void` `solveQueries(``int` `arr[], ``int` `n, ` `                         ``int` `q[], ``int` `m) ` `{ ` `    ``even = ``0``; ` `    ``odd = ``0``; ` `    ``keep_count(arr, n); ` ` `  `    ``for``(``int` `i = ``0``; i < m; i++) ` `    ``{ ` `       ``int` `X = q[i]; ` `        `  `       ``// Store set bits in X ` `       ``int` `count = ``0``; ` `        `  `       ``// Count set bits of X ` `       ``while` `(X != ``0``)  ` `       ``{ ` `           ``X = (X - ``1``) & X; ` `           ``count++; ` `       ``} ` `       ``if` `(count % ``2` `== ``0``)  ` `       ``{ ` `           ``System.out.print(even + ``" "` `+  ` `                             ``odd + ``"\n"``); ` `       ``} ` `       ``else` `       ``{ ` `           ``System.out.print(odd + ``" "` `+  ` `                           ``even + ``"\n"``); ` `       ``} ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``7``, ``4``, ``5``, ``3` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``int` `q[] = { ``3``, ``4``, ``12``, ``6` `}; ` `    ``int` `m = q.length; ` ` `  `    ``solveQueries(arr, n, q, m); ` `} ` `} ` ` `  `// This code is contributed by amal kumar choubey `

## Python3

 `# Python3 program to count number ` `# of even and odd set bits ` `# elements after XOR with a ` `# given element ` ` `  `even ``=` `0` `odd ``=` `0` ` `  `def` `keep_count(arr, N): ` `     `  `    ``global` `even ` `    ``global` `odd ` `     `  `    ``# Store the count of set bits ` `    ``for` `i ``in` `range``(N): ` `        ``count ``=` `0` ` `  `        ``# Brian Kernighan's algorithm ` `        ``while` `(arr[i] !``=` `0``): ` `            ``arr[i] ``=` `(arr[i] ``-` `1``) & arr[i] ` `            ``count ``+``=` `1` ` `  `        ``if` `(count ``%` `2` `=``=` `0``): ` `            ``even ``+``=` `1` `        ``else``: ` `            ``odd ``+``=` `1` ` `  `    ``return` ` `  `# Function to solve Q queries ` `def` `solveQueries(arr, n, q, m): ` `     `  `    ``global` `even ` `    ``global` `odd ` `     `  `    ``keep_count(arr, n) ` ` `  `    ``for` `i ``in` `range``(m): ` `        ``X ``=` `q[i] ` ` `  `        ``# Store set bits in X ` `        ``count ``=` `0` ` `  `        ``# Count set bits of X ` `        ``while` `(X !``=` `0``): ` `            ``X ``=` `(X ``-` `1``) & X ` `            ``count ``+``=` `1` ` `  `        ``if` `(count ``%` `2` `=``=` `0``): ` `            ``print``(even, odd) ` `        ``else``: ` `            ``print``(odd, even) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``arr ``=` `[ ``2``, ``7``, ``4``, ``5``, ``3` `] ` `    ``n ``=` `len``(arr) ` ` `  `    ``q ``=` `[ ``3``, ``4``, ``12``, ``6` `] ` `    ``m ``=` `len``(q) ` ` `  `    ``solveQueries(arr, n, q, m) ` ` `  `# This code is contributed by samarth `

## C#

 `// C# program to count number ` `// of even and odd set bits ` `// elements after XOR with a ` `// given element ` `using` `System; ` `class` `GFG{ ` `     `  `static` `int` `even, odd; ` ` `  `static` `void` `keep_count(``int` `[]arr, ``int` `N) ` `{ ` `     `  `    ``// Store the count of set bits ` `    ``int` `count; ` `     `  `    ``for``(``int` `i = 0; i < N; i++) ` `    ``{ ` `        ``count = 0; ` `             `  `        ``// Brian Kernighan's algorithm ` `        ``while` `(arr[i] != 0) ` `        ``{ ` `            ``arr[i] = (arr[i] - 1) & arr[i]; ` `            ``count++; ` `        ``} ` `        ``if` `(count % 2 == 0) ` `            ``even++; ` `        ``else` `            ``odd++; ` `    ``} ` `    ``return``; ` `} ` ` `  `// Function to solve Q queries ` `static` `void` `solveQueries(``int` `[]arr, ``int` `n, ` `                         ``int` `[]q, ``int` `m) ` `{ ` `    ``even = 0; ` `    ``odd = 0; ` `    ``keep_count(arr, n); ` ` `  `    ``for``(``int` `i = 0; i < m; i++) ` `    ``{ ` `        ``int` `X = q[i]; ` `             `  `        ``// Store set bits in X ` `        ``int` `count = 0; ` `             `  `        ``// Count set bits of X ` `        ``while` `(X != 0)  ` `        ``{ ` `            ``X = (X - 1) & X; ` `            ``count++; ` `        ``} ` `        ``if` `(count % 2 == 0)  ` `        ``{ ` `            ``Console.Write(even + ``" "` `+  ` `                           ``odd + ``"\n"``); ` `        ``} ` `        ``else` `        ``{ ` `            ``Console.Write(odd + ``" "` `+  ` `                         ``even + ``"\n"``); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 2, 7, 4, 5, 3 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``int` `[]q = { 3, 4, 12, 6 }; ` `    ``int` `m = q.Length; ` ` `  `    ``solveQueries(arr, n, q, m); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```2 3
3 2
2 3
2 3
```

Time Complexity: O(N * log N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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