Given a number N, the task is to find the count of X such that N XOR X == N OR X, where 0<=X<=N
Examples:
Input: N = 5
Output: 2
For N = 5,
5 XOR 2 == 5 OR 2
5 XOR 0 == 5 OR 0
Thus, count is 2.
Input: N = 7
Output: 1
For N = 7,
7 XOR 0 == 7 OR 0
Thus, count is 1.
Approach: The idea is to convert a given number to binary and then count the unset bits in it. 2^count gives us the number of X such that N XOR X == N OR X.
Below is the implementation of the above approach:
C++
// C++ program to find // the XOR equals OR count #include<iostream>#include<math.h>using namespace std; class gfg { // Function to calculate count // of numbers with XOR equals OR public: int xorEqualsOrCount(int N) { // variable to store count of unset bits int count = 0; int bit; while (N > 0) { bit = N % 2; if (bit == 0) count++; N = N / 2; } return (int)pow(2, count); } }; // Driver code int main() { gfg g ; int N = 7; cout<<g.xorEqualsOrCount(N); return 0; } // This code is contributed by Soumik |
Java
// Java program to find the XOR equals OR countimport java.io.*;import java.util.*;class GFG { // Function to calculate count of numbers with XOR equals OR static int xorEqualsOrCount(int N) { // variable to store count of unset bits int count = 0; int bit; while (N > 0) { bit = N % 2; if (bit == 0) count++; N = N / 2; } return (int)Math.pow(2, count); } // Driver code public static void main(String args[]) { int N = 7; System.out.println(xorEqualsOrCount(N)); }} |
Python3
# Python3 program to find# the XOR equals OR count# Function to calculate count# of numbers with XOR equals ORdef xorEqualsOrCount(N) : # variable to store # count of unset bits count = 0 while(N > 0) : bit = N % 2 if bit == 0 : count += 1 N //= 2 return int(pow(2, count))# Driver code if __name__ == "__main__" : N = 7 print(xorEqualsOrCount(N)) # This code is contributed by # ANKITRAI1 |
C#
// C# program to find // the XOR equals OR countusing System;class GFG { // Function to calculate count // of numbers with XOR equals OR static int xorEqualsOrCount(int N) { // variable to store count of unset bits int count = 0; int bit; while (N > 0) { bit = N % 2; if (bit == 0) count++; N = N / 2; } return (int)Math.Pow(2, count); } // Driver code public static void Main() { int N = 7; Console.WriteLine(xorEqualsOrCount(N)); }}// This code is contributed by inder_verma.. |
PHP
<?php // PHP program to find the XOR// equals OR count // Function to calculate count // of numbers with XOR equals OR function xorEqualsOrCount($N) { // variable to store count // of unset bits $count = 0; while ($N > 0) { $bit = $N % 2; if ($bit == 0) $count++; $N = intval($N / 2); } return pow(2, $count); } // Driver code $N = 7; echo xorEqualsOrCount($N);// This code is contributed// by ChitraNayal?> |
Javascript
<script>// Javascript program to find // the XOR equals OR count // Function to calculate count // of numbers with XOR equals OR function xorEqualsOrCount(N) { // variable to store count of unset bits let count = 0; let bit; while (N > 0) { bit = N % 2; if (bit == 0) count++; N = parseInt(N / 2); } return Math.pow(2, count); } // Driver code let N = 7; document.write(xorEqualsOrCount(N));</script> |
Output:
1
Time Complexity: O(log N)
Auxiliary Space: O(1)
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