Given two integer A and B, the task is to check whether the binary representation of B can be generated by permuting the binary digits of A.
Examples:
Input: A = 3, B = 9
Output: Yes
Binary(3) = 0011 and Binary(9) = 1001Input: A = 6, B = 7
Output: No
Approach: The idea is to count the number of set bits in the binary representations of both the numbers, now if they are equal then the answer is Yes or else the answer is No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if the // binary representation of b can be // generated by permuting the // binary digits of a bool isPossible(int a, int b) { // Find the count of set bits // in both the integers int cntA = __builtin_popcount(a); int cntB = __builtin_popcount(b); // If both the integers have // equal count of set bits if (cntA == cntB) return true; return false; } // Driver code int main() { int a = 3, b = 9; if (isPossible(a, b)) cout << "Yes"; else cout << "No"; return 0; } |
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Java
// Java implementation of the approach class GFG { // recursive function to count set bits public static int countSetBits(int n) { // base case if (n == 0) return 0; else // if last bit set add 1 else add 0 return (n & 1) + countSetBits(n >> 1); } // Function that returns true if the // binary representation of b can be // generated by permuting the // binary digits of a static boolean isPossible(int a, int b) { // Find the count of set bits // in both the integers int cntA = countSetBits(a); int cntB = countSetBits(b); // If both the integers have // equal count of set bits if (cntA == cntB) return true; return false; } // Driver code public static void main (String[] args) { int a = 3, b = 9; if (isPossible(a, b)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 implementation of the approach # function to count set bits def bitsoncount(x): return bin(x).count('1') # Function that returns true if the # binary representation of b can be # generated by permuting the # binary digits of a def isPossible(a, b): # Find the count of set bits # in both the integers cntA = bitsoncount(a); cntB = bitsoncount(b); # If both the integers have # equal count of set bits if (cntA == cntB): return True return False # Driver code a = 3b = 9 if (isPossible(a, b)): print("Yes") else: print("No") # This code is contributed by Sanjit Prasad |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { // recursive function to count set bits public static int countSetBits(int n) { // base case if (n == 0) return 0; else // if last bit set.Add 1 else.Add 0 return (n & 1) + countSetBits(n >> 1); } // Function that returns true if the // binary representation of b can be // generated by permuting the // binary digits of a static bool isPossible(int a, int b) { // Find the count of set bits // in both the integers int cntA = countSetBits(a); int cntB = countSetBits(b); // If both the integers have // equal count of set bits if (cntA == cntB) return true; return false; } // Driver code public static void Main (String[] args) { int a = 3, b = 9; if (isPossible(a, b)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by Rajput-Ji |
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Output:
Yes
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