Open In App
Related Articles

Toggle bits in the given range

Improve Article
Improve
Save Article
Save
Like Article
Like

Given a non-negative number n and two values l and r. The problem is to toggle the bits in the range l to r in the binary representation of n, i.e, to toggle bits from the rightmost lth bit to the rightmost rth bit. A toggle operation flips a bit 0 to 1 and a bit 1 to 0.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Examples: 

Input: n = 17, l = 2, r = 3
Output: 23
Explanation: (17)10 = (10001)2
                       (23)10 = (10111)2
The bits in the range 2 to 3 in the binary representation of 17 are toggled.

Input: n = 50, l = 2, r = 5
Output: 44

Recommended Practice

Approach: Following are the steps:

  1. Calculate num as = ((1 << r) – 1) ^ ((1 << (l-1)) – 1) or as ((1 <<r)-l). This will produce a number num having r number of bits and bits in the range l to r are the only set bits.
  2. Now, perform n = n ^ num. This will toggle the bits in the range l to r in n.

C++




// C++ implementation to toggle bits in
// the given range
#include <bits/stdc++.h>
using namespace std;
 
// function to toggle bits in the given range
unsigned int toggleBitsFromLToR(unsigned int n,
                                unsigned int l,
                                unsigned int r)
{
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
    // toggle bits in the range l to r in 'n'
    // and return the number
    // Besides this, we can calculate num as: num=(1<<r)-l .
    return (n ^ num);
}
 
// Driver program to test above
int main()
{
    unsigned int n = 50;
    unsigned int l = 2, r = 5;
    cout << toggleBitsFromLToR(n, l, r);
    return 0;
}


Java




// Java implementation to toggle bits in
// the given range
import java.io.*;
 
class GFG {
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
        // toggle bits in the range l to r in 'n'
        // and return the number
        // Besides this, we can calculate num as:
        // num=(1<<r)-l .
        return (n ^ num);
    }
 
    // driver program
    public static void main(String[] args)
    {
        int n = 50;
        int l = 2, r = 5;
        System.out.println(toggleBitsFromLToR(n, l, r));
    }
}
 
// Contributed by Pramod Kumar


Python3




# Python implementation
# to toggle bits in
# the given range
 
# function to toggle bits
# in the given range
 
 
def toggleBitsFromLToR(n, l, r):
 
    # calculating a number
    # 'num' having 'r'
    # number of bits and
    # bits in the range l
    # to r are the only set bits
    num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)
 
    # toggle bits in the
    # range l to r in 'n'
    # Besides this, we can calculate num as: num=(1<<r)-l .
 
    # and return the number
    return (n ^ num)
 
# Driver code
 
 
n = 50
l = 2
r = 5
 
print(toggleBitsFromLToR(n, l, r))
 
# This code is contributed
# by Anant Agarwal.


C#




// C# implementation to toggle bits
// in the given range
using System;
 
namespace Toggle {
public class GFG {
 
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
        // toggle bits in the range l to r in 'n'
        // Besides this, we can calculate num as:
        // num=(1<<r)-l .
        // and return the number
        return (n ^ num);
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 50;
        int l = 2, r = 5;
        Console.Write(toggleBitsFromLToR(n, l, r));
    }
}
}
 
// This code is contributed by Sam007.


PHP




<?php
// PHP implementation
// to toggle bits in
// the given range
 
// function to toggle bits
// in the given range
function toggleBitsFromLToR($n, $l, $r)
{
     
    // calculating a number
    // 'num' having 'r'
    // number of bits and
    // bits in the range l
    // to r are the only
    // set bits
    $num = ((1 << $r) - 1) ^
           ((1 << ($l - 1)) - 1);
 
    // toggle bits in the
    // range l to r in 'n'
    //Besides this, we can calculate num as: $num=(1<<$r)-$l .
    // and return the number
 
    return ($n ^ $num);
}
 
    // Driver Code
    $n = 50;
    $l = 2; $r = 5;
    echo toggleBitsFromLToR($n, $l, $r);
 
// This code is contributed by anuj_67
?>


Javascript




<script>
 
// Javascript implementation to toggle bits in
// the given range
 
// function to toggle bits in the given range
function toggleBitsFromLToR(n, l, r)
{
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    var num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
    // toggle bits in the range l to r in 'n'
    // and return the number
//Besides this, we can calculate num as: num=(1<<r)-l .
    return (n ^ num);
}
 
// Driver program to test above
var n = 50;
var l = 2, r = 5;
document.write( toggleBitsFromLToR(n, l, r));
 
</script>


Output

44

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

Approach 2: 
Iterate over the given range from L to R and check if the ith bit is set or not. if the ith bit is set then make it unset otherwise make it set bit.

C++




// C++ implementation to toggle bits in
// the given range
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to toggle bits in the given range
int toggleBitsFromLToR(int N, int L, int R)
{
    int res = N;
    for (int i = L; i <= R; i++) {
 
        // Set bit
        if ((N & (1 << (i - 1))) != 0) {
 
            // XOR will set 0 to already set
            // bits(a^a=0)
            res = res ^ (1 << (i - 1));
        }
 
        // unset bits
        else {
            // OR will set'0'bits to 1
            res = res | (1 << (i - 1));
        }
    }
    return res;
}
 
// Driver code
int main()
{
    int n = 50;
    int l = 2, r = 5;
    cout << toggleBitsFromLToR(n, l, r);
 
    return 0;
}
 
// This code is contributed by phasing17


Java




// Java implementation to toggle bits in
// the given range
import java.io.*;
 
class GFG {
 
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int N, int L, int R)
    {
        int res = N;
        for (int i = L; i <= R; i++) {
 
            // Set bit
            if ((N & (1 << (i - 1))) != 0) {
 
                // XOR will set 0 to already set
                // bits(a^a=0)
                res = res ^ (1 << (i - 1));
            }
 
            // unset bits
            else {
                // OR will set'0'bits to 1
                res = res | (1 << (i - 1));
            }
        }
        return res;
    }
 
    // Driver method
    public static void main(String[] args)
    {
        int n = 50;
        int l = 2, r = 5;
        System.out.println(toggleBitsFromLToR(n, l, r));
    }
}
 
// Contributed by Ocean Bhardwaj


Python3




# Python3 implementation to toggle bits in
# the given range
 
# Function to toggle bits in the given range
def toggleBitsFromLToR(N, L, R):
 
    res = N
    for i in range(L, R + 1):
 
        # Set bit
        if ((N & (1 << (i - 1))) != 0):
 
            # XOR will set 0 to already set
            # bits(a^a=0)
            res = res ^ (1 << (i - 1))
 
        # unset bits
        else:
            # OR will set'0'bits to 1
            res = res | (1 << (i - 1))
 
    return res
 
# Driver code
n = 50
l = 2
r = 5
print(toggleBitsFromLToR(n, l, r))
 
# This code is contributed by phasing17


C#




// C# implementation to toggle bits in
// the given range
using System;
 
class GFG {
 
  // Function to toggle bits in the given range
  static int toggleBitsFromLToR(int N, int L, int R)
  {
    int res = N;
    for (int i = L; i <= R; i++) {
 
      // Set bit
      if ((N & (1 << (i - 1))) != 0) {
 
        // XOR will set 0 to already set
        // bits(a^a=0)
        res = res ^ (1 << (i - 1));
      }
 
      // unset bits
      else {
        // OR will set'0'bits to 1
        res = res | (1 << (i - 1));
      }
    }
    return res;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int n = 50;
    int l = 2, r = 5;
 
    // Function call
    Console.WriteLine(toggleBitsFromLToR(n, l, r));
  }
}
 
// This code is Contributed by phasing17


Javascript




// JavaScript implementation to toggle bits in
// the given range
 
// Function to toggle bits in the given range
function toggleBitsFromLToR(N, L, R)
{
    let res = N;
    for (let i = L; i <= R; i++) {
 
        // Set bit
        if ((N & (1 << (i - 1))) != 0) {
 
            // XOR will set 0 to already set
            // bits(a^a=0)
            res = res ^ (1 << (i - 1));
        }
 
        // unset bits
        else {
            // OR will set'0'bits to 1
            res = res | (1 << (i - 1));
        }
    }
    return res;
}
 
 
// Driver code
let n = 50;
let l = 2, r = 5;
console.log(toggleBitsFromLToR(n, l, r));
 
 
 
// This code is contributed by phasing17


Output

44

Time Complexity: O(R – L + 1) 
Auxiliary Space: O(1)

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 08 Aug, 2022
Like Article
Save Article
Similar Reads
Related Tutorials