Unset least significant K bits of a given number
Given an integer N, the task is to print the number obtained by unsetting the least significant K bits from N.
Examples:
Input: N = 200, K=5
Output: 192
Explanation:
(200)10 = (11001000)2
Unsetting least significant K(= 5) bits from the above binary representation, the new number obtained is (11000000)2 = (192)10Input: N = 730, K = 3
Output: 720
Approach: Follow the steps below to solve the problem:
- The idea is to create a mask of the form 111111100000….
- To create a mask, start from all ones as 1111111111….
- There are two possible options to generate all 1s. Either generate it by flipping all 0s with 1s or by using 2s complement and left shift it by K bits.
mask = ((~0) << K + 1) or
mask = (-1 << K + 1)
- Finally, print the value of K + 1 as it is zero-based indexing from the right to left.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to return the value // after unsetting K LSBs int clearLastBit( int N, int K) { // Create a mask int mask = (-1 << K + 1); // Bitwise AND operation with // the number and the mask return N = N & mask; } // Driver Code int main() { // Given N and K int N = 730, K = 3; // Function Call cout << clearLastBit(N, K); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to return the value // after unsetting K LSBs static int clearLastBit( int N, int K) { // Create a mask int mask = (- 1 << K + 1 ); // Bitwise AND operation with // the number and the mask return N = N & mask; } // Driver Code public static void main(String[] args) { // Given N and K int N = 730 , K = 3 ; // Function Call System.out.print(clearLastBit(N, K)); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach # Function to return the value # after unsetting K LSBs def clearLastBit(N, K): # Create a mask mask = ( - 1 << K + 1 ) # Bitwise AND operation with # the number and the mask N = N & mask return N # Driver Code # Given N and K N = 730 K = 3 # Function call print (clearLastBit(N, K)) # This code is contributed by Shivam Singh |
C#
// C# program for the above approach using System; class GFG{ // Function to return the value // after unsetting K LSBs static int clearLastBit( int N, int K) { // Create a mask int mask = (-1 << K + 1); // Bitwise AND operation with // the number and the mask return N = N & mask; } // Driver Code public static void Main(String[] args) { // Given N and K int N = 730, K = 3; // Function Call Console.Write(clearLastBit(N, K)); } } // This code is contributed by shikhasingrajput |
Javascript
<script> // javascript program for the above approach // Function to return the value // after unsetting K LSBs function clearLastBit(N , K) { // Create a mask var mask = (-1 << K + 1); // Bitwise AND operation with // the number and the mask return N = N & mask; } // Driver Code //Given N and K var N = 730, K = 3; // Function Call document.write(clearLastBit(N, K)); // This code contributed by shikhasingrajput </script> |
Output:
720
Time Complexity: O(1)
Auxiliary Space: O(1)