Given a number X, calculate number of possible pairs (a, b) such that bitwise or of a and b is equal to X and number of bits in both a and b is less than equal to number of bits in X.
Examples:
Input: X = 6
Output: 9
Explanation:
The possible pairs of (a, b) are (4, 6), (6, 4), (6, 6), (6, 2), (4, 2), (6, 0), (2, 6), (2, 4), (0, 6).Input: X = 21
Output: 27
Explanation:
In total there are 27 pairs possible.
Approach: To solve the problem mentioned above follow the steps given below:
- Iterate through every bit of given number X.
- If the bit is 1 then from the truth table of Bitwise OR we know that there are 3 combinations possible for that given bit in number a and b that is (0, 1), (1, 0), (1, 1) that is 3 possible ways.
- If the bit is 0 then from the truth table of Bitwise OR we know that there is only 1 combination possible for that given bit in number a and b that is (0, 0).
- So our answer will be answer will be 3 ^ (number of on bits in X).
Below is the implementation of above approach:
C++
// C++ implementation to Count number of// possible pairs of (a, b) such that// their Bitwise OR gives the value X#include <iostream>using namespace std;// Function to count the pairsint count_pairs(int x){ // Initializing answer with 1 int ans = 1; // Iterating through bits of x while (x > 0) { // check if bit is 1 if (x % 2 == 1) // multiplying ans by 3 // if bit is 1 ans = ans * 3; x = x / 2; } return ans;}// Driver codeint main(){ int X = 6; cout << count_pairs(X) << endl; return 0;} |
Java
// Java implementation to count number of// possible pairs of (a, b) such that// their Bitwise OR gives the value Xclass GFG{// Function to count the pairsstatic int count_pairs(int x){ // Initializing answer with 1 int ans = 1; // Iterating through bits of x while (x > 0) { // Check if bit is 1 if (x % 2 == 1) // Multiplying ans by 3 // if bit is 1 ans = ans * 3; x = x / 2; } return ans;}// Driver codepublic static void main(String[] args){ int X = 6; System.out.print(count_pairs(X) + "\n");}}// This code is contributed by amal kumar choubey |
Python3
# Python3 implementation to count number of# possible pairs of (a, b) such that# their Bitwise OR gives the value X# Function to count the pairsdef count_pairs(x): # Initializing answer with 1 ans = 1; # Iterating through bits of x while (x > 0): # Check if bit is 1 if (x % 2 == 1): # Multiplying ans by 3 # if bit is 1 ans = ans * 3; x = x // 2; return ans;# Driver codeif __name__ == '__main__': X = 6; print(count_pairs(X));# This code is contributed by amal kumar choubey |
C#
// C# implementation to count number of// possible pairs of (a, b) such that// their Bitwise OR gives the value Xusing System;class GFG{ // Function to count the pairs static int count_pairs(int x) { // Initializing answer with 1 int ans = 1; // Iterating through bits of x while (x > 0) { // Check if bit is 1 if (x % 2 == 1) // Multiplying ans by 3 // if bit is 1 ans = ans * 3; x = x / 2; } return ans; } // Driver code public static void Main(String[] args) { int X = 6; Console.Write(count_pairs(X) + "\n"); }}// This code is contributed by sapnasingh4991 |
Javascript
<script>// javascript implementation to count number of// possible pairs of (a, b) such that// their Bitwise OR gives the value X // Function to count the pairs function count_pairs(x) { // Initializing answer with 1 var ans = 1; // Iterating through bits of x while (x > 0) { // Check if bit is 1 if (x % 2 == 1) // Multiplying ans by 3 // if bit is 1 ans = ans * 3; x = parseInt(x / 2); } return ans; } // Driver code var X = 6; document.write(count_pairs(X) + "\n");// This code contributed by Rajput-Ji</script> |
Output:
9
Time complexity: O(log(X))
Auxiliary Space: O(1)
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