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)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:

720

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