Unset least significant K bits of a given number

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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#

// 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:

Time Complexity: O(1)
Auxiliary Space: O(1)

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