Toggle bits in the given range (original) (raw)

Last Updated : 1 Apr, 2024

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 **lth least significant bit bit to the **rth least significant bit (the rightmost bit as counted as first 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 = 1, r = 3
Output: 22
Explanation: (17) 10 = (10001)2
(22) 10 = (10110)2
The bits in the range **1 to **3 in the binary representation of **17 are toggled.

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

Explanation: (50) 10 = (110010)2
(44) 10 = (1****01100)2
The bits in the range **2 to **5 in the binary representation of **50 are toggled.

Try It Yourself

**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 (from rightmost end in binary representation) 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 = 17; unsigned int l = 1, r = 3; 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

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.

JavaScript

PHP

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.

`

**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

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

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

`

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