Clear Bit Ranges in Given Number in C++

Given a number n, write C++ program to clear bits in the given range between l and r. Where, 1 <= l <= r <= number of bits in the binary representation of n.

Approaches to clear bits in the given range

Following are the different approaches to clear bit ranges in given number in C++ ?

• Using the Brute Force Approach
• Using Optimized Approach

Using Brute Force Approach

Following are the steps to clear bit ranges in given number using Brute force approach ?

• Step 1: Create a function to clear the i-th bit named clearIthBit().
  • Create a mask with 0 at the i-th position and 1s everywhere else.
  • Apply the mask to n using bitwise AND to clear the i-th bit.
• Step 2: Use a loop to clear each bit from index l to r individually and call the function clearIthBit().

Example

#include <iostream>

using namespace std;

int clearIthBit(int n, int i) {
  int mask = ~(1 << i);
  return n & mask;
}

int main() {
  int n = 63;
  int l = 1, r = 3;
  int result = 0;

  for (int i = l - 1; i < r; i++) {
    n = clearIthBit(n, i);
  }

  cout << n;

  return 0;
}

Space Complexity: O(1)
Time Complexity: O(r-l+1)  
Since we are running a loop from l to r.

Using Optimized Approach

Following are the steps to clear bit ranges in a given number using Optimized Approach?

• Step 1: Create a mask with all bits set to 1. This can be done by taking bitwise NOT(~) of 0. Since 0 can be 000....0000, its negation will be 111....1111.
• Step 2: Create a mask for all the right side bits i.e. we need to shift all 1's r times.
• Step 3: Create a mask for all left side bits i.e. we need to keep all bits 1 (l-1) times.
• Step 4: Combine masks to get a mask with 0s from l to r and 1s elsewhere.
• Step 5: Clear the bits from l to r in n using the mask.

Example

#include <iostream>
using namespace std;

int clearBits(int n, int l, int r) {
    int allOnes = ~0;
    int leftMask = allOnes << (r);
    int rightMask = (1 << (l-1)) - 1;
    int mask = leftMask | rightMask;
    return n & mask;
}

int main() {
    int n, l, r;
    cout << "Enter the number (n): ";
    cin >> n;
    cout << "Enter the left index (l): ";
    cin >> l;
    cout << "Enter the right index (r): ";
    cin >> r;

    int result = clearBits(n, l, r);
    cout << "Result after clearing bits from " << l << " to " << r << " is: " << result << endl;

    return 0;
}

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