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 3Input: arr[] = { 7, 1, 6, 5, 11 }, Q[] = { 2, 10, 3, 6 }
Output: 3 2
2 3
2 3
2 3
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 odd 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 <bits/stdc++.h>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 queriesvoid 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 codeint main(){ int arr[] = { 2, 7, 4, 5, 3 }; int n = sizeof(arr) / sizeof(arr[0]); int q[] = { 3, 4, 12, 6 }; int m = sizeof(q) / sizeof(q[0]); 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 elementclass 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 queriesstatic 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 codepublic 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 elementeven = 0odd = 0def 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 queriesdef 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 codeif __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 elementusing 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 queriesstatic 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 codepublic 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 |
Javascript
<script>// Javascript program to count number// of even and odd set bits// elements after XOR with a// given element function even, odd;function keep_count(arr, N){ // Store the count of set bits var count; for(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 queriesfunction solveQueries(arr, n, q, m) { even = 0; odd = 0; keep_count(arr, n); for(i = 0; i < m; i++) { var X = q[i]; // Store set bits in X var count = 0; // Count set bits of X while (X != 0) { X = (X - 1) & X; count++; } if (count % 2 == 0) { document.write(even + " " + odd + "<br/>"); } else { document.write(odd + " " + even + "<br/>"); } }}// Driver codevar arr = [ 2, 7, 4, 5, 3 ];var n = arr.length;var q = [ 3, 4, 12, 6 ];var m = q.length;solveQueries(arr, n, q, m);// This code is contributed by aashish1995</script> |
Output:
2 3 3 2 2 3 2 3
Time Complexity: O(N * log(max(arr))+m*log(max(q))) where max(arr) and max(q) represents the maximum element in array arr and q respectively.
Auxiliary Space: O(1) as constant space is used
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
