Given an integer X > 1 and an integer K > 0, the task is to find the greatest odd number < X such that the number of 1’s in its binary representation is at most K.
Examples:
Input: X = 10, K = 2
Output: 10Input: X = 29, K = 2
Output: 24
Naive Approach: Starting from X – 1 check all the numbers below X which have at most K set bits, the first number satisfying the condition is the required answer.
Efficient Approach: is to count the set bits. If the count is less than or equal to K, return X. Otherwise, keep removing rightmost set bit while count – k does not become 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the greatest number <= X// having at most K set bits.int greatestKBits(int X, int K){ int set_bit_count = __builtin_popcount(X); if (set_bit_count <= K) return X; // Remove rightmost set bits one // by one until we count becomes k int diff = set_bit_count - K; for (int i = 0; i < diff; i++) X &= (X - 1); // Return the required number return X;}// Driver codeint main(){ int X = 21, K = 2; cout << greatestKBits(X, K); return 0;} |
Java
// Java implementation of the approachimport java.io.*; class GFG { // Function to return the greatest number <= X // having at most K set bits. int greatestKBits(int X, int K) { int set_bit_count = Integer.bitCount(X); if (set_bit_count <= K) return X; // Remove rightmost set bits one // by one until we count becomes k int diff = set_bit_count - K; for (int i = 0; i < diff; i++) X &= (X - 1); // Return the required number return X; }// Driver codepublic static void main (String[] args) { int X = 21, K = 2; GFG g=new GFG(); System.out.print(g.greatestKBits(X, K));}//This code is contributed by Shivi_Aggarwal} |
Python3
# Python 3 implementation of the approach# Function to return the greatest # number <= X having at most K set bits.def greatestKBits(X, K): set_bit_count = bin(X).count('1') if (set_bit_count <= K): return X # Remove rightmost set bits one # by one until we count becomes k diff = set_bit_count - K for i in range(0, diff, 1): X &= (X - 1) # Return the required number return X# Driver codeif __name__ == '__main__': X = 21 K = 2 print(greatestKBits(X, K)) # This code is contributed by# Shashank_Sharma |
C#
// C# implementation of the above approach using System;class GFG{ // Function to get no of set // bits in binary representation // of positive integer n static int countSetBits(int n) { int count = 0; while (n > 0) { count += n & 1; n >>= 1; } return count; } // Function to return the greatest number <= X // having at most K set bits. static int greatestKBits(int X, int K) { int set_bit_count = countSetBits(X); if (set_bit_count <= K) return X; // Remove rightmost set bits one // by one until we count becomes k int diff = set_bit_count - K; for (int i = 0; i < diff; i++) X &= (X - 1); // Return the required number return X; } // Driver code public static void Main() { int X = 21, K = 2; Console.WriteLine(greatestKBits(X, K)); } } // This code is contributed by Ryuga |
Javascript
<script>// Javascript implementation of the above approach// Function to get no of set// bits in binary representation// of positive integer nfunction countSetBits( n){ let count = 0; while (n > 0) { count += n & 1; n >>= 1; } return count;} // Function to return the greatest number <= X// having at most K set bits.function greatestKBits( X, K){ let set_bit_count = countSetBits(X); if (set_bit_count <= K) return X; // Remove rightmost set bits one // by one until we count becomes k let diff = set_bit_count - K; for (let i = 0; i < diff; i++) X &= (X - 1); // Return the required number return X;}// Driver Codelet X = 21, K = 2;document.write(greatestKBits(X, K));</script> |
Output
20
Time Complexity: O(log2X), as the time complexity of __builtin_popcount(X) is O(log2X)
Auxiliary Space: O(1)
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