Unset least significant K bits of a given number

Last Updated : 08 Apr, 2021

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)₁₀

Input: 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# 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>