Given two arrays **A[] and B[]** having **N** and **M** positive elements respectively. The task is to count the number of elements in array A with even number of set bits in XOR for every element of array B.

**Examples:**

Input:A[] = { 4, 2, 15, 9, 8, 8 }, B[] = { 3, 4, 22 }

Output:2 4 4

Explanation:

Binary representation of elements of A are : 100, 10, 1111, 1001, 1000, 1000

Binary representation of elements of B are : 11, 101, 10110

Now for element 3(11),

3^4 = 11^100 = 111

3^2 = 11^10 = 01

3^15 = 11^1111 = 1100

3^9 = 11^1001 = 1111

3^8 = 11^1000 = 1011

3^8 = 11^1000 = 1011

Only 2 elements {15, 9} in A[] are there for element 3 such that count of set bit after XOR is even. So the count is 2.

Similarly, Count for element 4 and 22 is 4.

Input:A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, B[] = { 4 }

Output:5

Explanation:

The element in A[] such that count of set bit after XOR is even is {1, 2, 4, 7, 8}. So the count is 5.

**Naive Approach:** The idea is to compute the XOR for every element in the array **B[]** with each element in the array **A[]** and count the number having even set bit.

**Time Complexity:** O(N*M), where N and M is the length of array **A[]** and **B[]** respectively.

**Efficient Approach:** The idea is to use the property of XOR. For any two numbers, if the count of set bit for both the numbers are even or odd then count of the set bit after XOR of both numbers is **even**.

Below are the steps based on the above property:

- Count the number of element in the array
**A[]**having even(say**a**) and odd(say**b**) number of set bits. - For each element in the array
**B[]**:- If current element have even count of set bit, then the number element in the array
**A[]**whose XOR with the current element has even count of set bit is**a**. - If current element have odd count of set bit, then the number element in the array
**A[]**whose XOR with the current element has even count of set bit is**b**.

- If current element have even count of set bit, then the number element in the array

Below is the implementation of the above approach:

## CPP

`// C++ program for the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function that count the XOR of B[] ` `// with all the element in A[] having ` `// even set bit ` `void` `countEvenBit(` `int` `A[], ` `int` `B[], ` `int` `n, ` `int` `m) ` `{ ` ` ` `int` `i, j, cntOdd = 0, cntEven = 0; ` ` ` `for` `(i = 0; i < n; i++) { ` ` ` ` ` `// Count the set bits in A[i] ` ` ` `int` `x = __builtin_popcount(A[i]); ` ` ` ` ` `// check for even or Odd ` ` ` `if` `(x & 1) { ` ` ` `cntEven++; ` ` ` `} ` ` ` `else` `{ ` ` ` `cntOdd++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// To store the count of element for ` ` ` `// B[] such that XOR with all the ` ` ` `// element in A[] having even set bit ` ` ` `int` `CountB[m]; ` ` ` ` ` `for` `(i = 0; i < m; i++) { ` ` ` ` ` `// Count set bit for B[i] ` ` ` `int` `x = __builtin_popcount(B[i]); ` ` ` ` ` `// check for Even or Odd ` ` ` `if` `(x & 1) { ` ` ` `CountB[i] = cntEven; ` ` ` `} ` ` ` `else` `{ ` ` ` `CountB[i] = cntOdd; ` ` ` `} ` ` ` `} ` ` ` ` ` `for` `(i = 0; i < m; i++) { ` ` ` `cout << CountB[i] << ` `' '` `; ` ` ` `} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `A[] = { 4, 2, 15, 9, 8, 8 }; ` ` ` `int` `B[] = { 3, 4, 22 }; ` ` ` ` ` `countEvenBit(A, B, 6, 3); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program for the above approach ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function that count the XOR of B[] ` `// with all the element in A[] having ` `// even set bit ` `static` `void` `countEvenBit(` `int` `A[], ` `int` `B[], ` `int` `n, ` `int` `m) ` `{ ` ` ` `int` `i, j, cntOdd = ` `0` `, cntEven = ` `0` `; ` ` ` `for` `(i = ` `0` `; i < n; i++) { ` ` ` ` ` `// Count the set bits in A[i] ` ` ` `int` `x = Integer.bitCount(A[i]); ` ` ` ` ` `// check for even or Odd ` ` ` `if` `(x % ` `2` `== ` `1` `) { ` ` ` `cntEven++; ` ` ` `} ` ` ` `else` `{ ` ` ` `cntOdd++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// To store the count of element for ` ` ` `// B[] such that XOR with all the ` ` ` `// element in A[] having even set bit ` ` ` `int` `[]CountB = ` `new` `int` `[m]; ` ` ` ` ` `for` `(i = ` `0` `; i < m; i++) { ` ` ` ` ` `// Count set bit for B[i] ` ` ` `int` `x = Integer.bitCount(B[i]); ` ` ` ` ` `// check for Even or Odd ` ` ` `if` `(x%` `2` `== ` `1` `) { ` ` ` `CountB[i] = cntEven; ` ` ` `} ` ` ` `else` `{ ` ` ` `CountB[i] = cntOdd; ` ` ` `} ` ` ` `} ` ` ` ` ` `for` `(i = ` `0` `; i < m; i++) { ` ` ` `System.out.print(CountB[i] +` `" "` `); ` ` ` `} ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `A[] = { ` `4` `, ` `2` `, ` `15` `, ` `9` `, ` `8` `, ` `8` `}; ` ` ` `int` `B[] = { ` `3` `, ` `4` `, ` `22` `}; ` ` ` ` ` `countEvenBit(A, B, ` `6` `, ` `3` `); ` `} ` `} ` ` ` `// This code is contributed by sapnasingh4991 ` |

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## Python3

`# Python3 program for the above approach ` ` ` `# Function that count the XOR of B ` `# with all the element in A having ` `# even set bit ` `def` `countEvenBit(A, B, n, m): ` ` ` ` ` `i, j, cntOdd ` `=` `0` `, ` `0` `, ` `0` ` ` `cntEven ` `=` `0` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# Count the set bits in A[i] ` ` ` `x ` `=` `bin` `(A[i])[` `2` `:].count(` `'1'` `) ` ` ` ` ` `# check for even or Odd ` ` ` `if` `(x & ` `1` `): ` ` ` `cntEven ` `+` `=` `1` ` ` ` ` `else` `: ` ` ` `cntOdd ` `+` `=` `1` ` ` ` ` `# To store the count of element for ` ` ` `# B such that XOR with all the ` ` ` `# element in A having even set bit ` ` ` `CountB ` `=` `[` `0` `]` `*` `m ` ` ` ` ` `for` `i ` `in` `range` `(m): ` ` ` ` ` `# Count set bit for B[i] ` ` ` `x ` `=` `bin` `(B[i])[` `2` `:].count(` `'1'` `) ` ` ` ` ` `# check for Even or Odd ` ` ` `if` `(x & ` `1` `): ` ` ` `CountB[i] ` `=` `cntEven ` ` ` ` ` `else` `: ` ` ` `CountB[i] ` `=` `cntOdd ` ` ` ` ` `for` `i ` `in` `range` `(m): ` ` ` `print` `(CountB[i], end` `=` `" "` `) ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `A ` `=` `[ ` `4` `, ` `2` `, ` `15` `, ` `9` `, ` `8` `, ` `8` `] ` ` ` `B ` `=` `[ ` `3` `, ` `4` `, ` `22` `] ` ` ` ` ` `countEvenBit(A, B, ` `6` `, ` `3` `) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# program for the above approach ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function that count the XOR of []B ` `// with all the element in []A having ` `// even set bit ` `static` `void` `countEvenBit(` `int` `[]A, ` `int` `[]B, ` `int` `n, ` `int` `m) ` `{ ` ` ` `int` `i, cntOdd = 0, cntEven = 0; ` ` ` `for` `(i = 0; i < n; i++) ` ` ` `{ ` ` ` ` ` `// Count the set bits in A[i] ` ` ` `int` `x = bitCount(A[i]); ` ` ` ` ` `// check for even or Odd ` ` ` `if` `(x % 2 == 1) { ` ` ` `cntEven++; ` ` ` `} ` ` ` `else` `{ ` ` ` `cntOdd++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// To store the count of element for ` ` ` `// []B such that XOR with all the ` ` ` `// element in []A having even set bit ` ` ` `int` `[]CountB = ` `new` `int` `[m]; ` ` ` ` ` `for` `(i = 0; i < m; i++) { ` ` ` ` ` `// Count set bit for B[i] ` ` ` `int` `x = bitCount(B[i]); ` ` ` ` ` `// check for Even or Odd ` ` ` `if` `(x % 2 == 1) { ` ` ` `CountB[i] = cntEven; ` ` ` `} ` ` ` `else` `{ ` ` ` `CountB[i] = cntOdd; ` ` ` `} ` ` ` `} ` ` ` ` ` `for` `(i = 0; i < m; i++) { ` ` ` `Console.Write(CountB[i] +` `" "` `); ` ` ` `} ` `} ` `static` `int` `bitCount(` `int` `x) ` `{ ` ` ` `int` `setBits = 0; ` ` ` `while` `(x != 0) { ` ` ` `x = x & (x - 1); ` ` ` `setBits++; ` ` ` `} ` ` ` `return` `setBits; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `[]A = { 4, 2, 15, 9, 8, 8 }; ` ` ` `int` `[]B = { 3, 4, 22 }; ` ` ` ` ` `countEvenBit(A, B, 6, 3); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

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**Output:**

2 4 4

**Time Complexity:** O(N + M), where N and M are the length of the given two array respectively.

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