Count of even set bits between XOR of two arrays
Last Updated :
18 Sep, 2023
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.
Approach:
- First we create a function countSetBits which returns the number of set bits in the binary representation of a given integer.
- Then we Iterate through the array B[] and for each element, iterate through the array A[] and count the number of elements in A[] with even number of set bits in XOR with the current element of B[].
- Print the count obtained in step 2 for each element of B[].
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int countSetBits( int n){
int count = 0;
while (n){
count += n & 1;
n >>= 1;
}
return count;
}
void countEvenSetBits( int A[], int B[], int N, int M){
for ( int i=0; i<M; i++){
int count = 0;
for ( int j=0; j<N; j++){
if (countSetBits(B[i] ^ A[j]) % 2 == 0){
count++;
}
}
cout << count << " " ;
}
}
int main() {
int A[] = { 4, 2, 15, 9, 8, 8 };
int B[] = { 3, 4, 22 };
int N= sizeof (A)/ sizeof (A[0]);
int M= sizeof (B)/ sizeof (B[0]);
countEvenSetBits(A, B,N,M);
return 0;
}
|
Java
public class CountEvenSetBits {
static int countSetBits( int n) {
int count = 0 ;
while (n != 0 ) {
count += n & 1 ;
n >>= 1 ;
}
return count;
}
static void countEvenSetBits( int [] A, int [] B, int N, int M) {
for ( int i = 0 ; i < M; i++) {
int count = 0 ;
for ( int j = 0 ; j < N; j++) {
if (countSetBits(B[i] ^ A[j]) % 2 == 0 ) {
count++;
}
}
System.out.print(count + " " );
}
}
public static void main(String[] args) {
int [] A = { 4 , 2 , 15 , 9 , 8 , 8 };
int [] B = { 3 , 4 , 22 };
int N = A.length;
int M = B.length;
countEvenSetBits(A, B, N, M);
}
}
|
Python3
def countSetBits(n):
count = 0
while n:
count + = n & 1
n >> = 1
return count
def countEvenSetBits(A, B, N, M):
for i in range (M):
count = 0
for j in range (N):
if countSetBits(B[i] ^ A[j]) % 2 = = 0 :
count + = 1
print (count, end = " " )
A = [ 4 , 2 , 15 , 9 , 8 , 8 ]
B = [ 3 , 4 , 22 ]
N = len (A)
M = len (B)
countEvenSetBits(A, B, N, M)
|
C#
using System;
class GFG
{
static int CountSetBits( int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
static void CountEvenSetBits( int [] A, int [] B, int N, int M)
{
for ( int i = 0; i < M; i++)
{
int count = 0;
for ( int j = 0; j < N; j++)
{
if (CountSetBits(B[i] ^ A[j]) % 2 == 0)
{
count++;
}
}
Console.Write(count + " " );
}
}
static void Main()
{
int [] A = { 4, 2, 15, 9, 8, 8 };
int [] B = { 3, 4, 22 };
int N = A.Length;
int M = B.Length;
CountEvenSetBits(A, B, N, M);
}
}
|
Javascript
function countSetBits(n) {
let count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
function countEvenSetBits(A, B) {
for (let i = 0; i < B.length; i++) {
let count = 0;
for (let j = 0; j < A.length; j++) {
if (countSetBits(B[i] ^ A[j]) % 2 === 0) {
count++;
}
}
console.log(count + " " );
}
}
let A = [4, 2, 15, 9, 8, 8];
let B = [3, 4, 22];
countEvenSetBits(A, B);
|
Time Complexity: O(N*M*log(max(A)+max(B)), where N and M is the length of array A[] and B[] respectively.
Auxiliary Space: O(1) as we are not using any extra space .
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.
Below is the implementation of the above approach:
CPP
#include <bits/stdc++.h>
using namespace std;
void countEvenBit( int A[], int B[], int n, int m)
{
int i, j, cntOdd = 0, cntEven = 0;
for (i = 0; i < n; i++) {
int x = __builtin_popcount(A[i]);
if (x & 1) {
cntEven++;
}
else {
cntOdd++;
}
}
int CountB[m];
for (i = 0; i < m; i++) {
int x = __builtin_popcount(B[i]);
if (x & 1) {
CountB[i] = cntEven;
}
else {
CountB[i] = cntOdd;
}
}
for (i = 0; i < m; i++) {
cout << CountB[i] << ' ' ;
}
}
int main()
{
int A[] = { 4, 2, 15, 9, 8, 8 };
int B[] = { 3, 4, 22 };
countEvenBit(A, B, 6, 3);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void countEvenBit( int A[], int B[], int n, int m)
{
int i, j, cntOdd = 0 , cntEven = 0 ;
for (i = 0 ; i < n; i++) {
int x = Integer.bitCount(A[i]);
if (x % 2 == 1 ) {
cntEven++;
}
else {
cntOdd++;
}
}
int []CountB = new int [m];
for (i = 0 ; i < m; i++) {
int x = Integer.bitCount(B[i]);
if (x% 2 == 1 ) {
CountB[i] = cntEven;
}
else {
CountB[i] = cntOdd;
}
}
for (i = 0 ; i < m; i++) {
System.out.print(CountB[i] + " " );
}
}
public static void main(String[] args)
{
int A[] = { 4 , 2 , 15 , 9 , 8 , 8 };
int B[] = { 3 , 4 , 22 };
countEvenBit(A, B, 6 , 3 );
}
}
|
Python3
def countEvenBit(A, B, n, m):
i, j, cntOdd = 0 , 0 , 0
cntEven = 0
for i in range (n):
x = bin (A[i])[ 2 :].count( '1' )
if (x & 1 ):
cntEven + = 1
else :
cntOdd + = 1
CountB = [ 0 ] * m
for i in range (m):
x = bin (B[i])[ 2 :].count( '1' )
if (x & 1 ):
CountB[i] = cntEven
else :
CountB[i] = cntOdd
for i in range (m):
print (CountB[i], end = " " )
if __name__ = = '__main__' :
A = [ 4 , 2 , 15 , 9 , 8 , 8 ]
B = [ 3 , 4 , 22 ]
countEvenBit(A, B, 6 , 3 )
|
C#
using System;
class GFG{
static void countEvenBit( int []A, int []B, int n, int m)
{
int i, cntOdd = 0, cntEven = 0;
for (i = 0; i < n; i++)
{
int x = bitCount(A[i]);
if (x % 2 == 1) {
cntEven++;
}
else {
cntOdd++;
}
}
int []CountB = new int [m];
for (i = 0; i < m; i++) {
int x = bitCount(B[i]);
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;
}
public static void Main(String[] args)
{
int []A = { 4, 2, 15, 9, 8, 8 };
int []B = { 3, 4, 22 };
countEvenBit(A, B, 6, 3);
}
}
|
Javascript
<script>
function countEvenBit(A, B, n, m)
{
let i, j, cntOdd = 0, cntEven = 0;
for (i = 0; i < n; i++) {
let x = bitCount(A[i]);
if (x & 1) {
cntEven++;
}
else {
cntOdd++;
}
}
let CountB = new Array(m);
for (i = 0; i < m; i++) {
let x = bitCount(B[i]);
if (x & 1) {
CountB[i] = cntEven;
}
else {
CountB[i] = cntOdd;
}
}
for (i = 0; i < m; i++) {
document.write(CountB[i] + " " );
}
}
function bitCount(x)
{
let setBits = 0;
while (x != 0) {
x = x & (x - 1);
setBits++;
}
return setBits;
}
let A = [ 4, 2, 15, 9, 8, 8 ];
let B = [ 3, 4, 22 ];
countEvenBit(A, B, 6, 3);
</script>
|
Time Complexity: O(N + M), where N and M are the length of the given two array respectively.
Auxiliary Space: O(M)
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