Number formed by flipping all bits to the left of rightmost set bit

Given an integer N, the task is to flip all the bits to the left of rightmost set bit and print the number generated.

Examples:

Input: N = 10 
Output: 6 
Explanation:
10 (1010 in binary) 
flipping all bits left to rightmost set bit (index 2) 
-> 6 (0110 in binary)

Input: N = 120 
Output: 8 
Explanation:
120 (1111000 in binary) 
flipping all bits left to rightmost set bit (index 3) 
-> 8 (0001000 in binary)

Naive Approach:
To solve the problem mentioned above we will follow the steps given below:

• Convert the given integer into its binary form and store each bit into an array.
• Traverse the array and break after the first occurrence of 1.
• Flip all bits to the left of that index. Convert that binary sequence into a decimal number and return it.

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

Efficient Approach:
To optimize the above method we will try to use Bitwise operators.

1. We are finding the set bit position from the rightmost side that is LSB and the total number of bits.
2. XOR the given number with the number which has all set bits having total set bits equal to the total number of bits.
3. We don’t want to flip the rightmost bits from a set bit. So we are again taking XOR with the number which has all set bits having total set bits equal to firstbit

Below is the implementation of the above approach:

C++

// C++ program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
#include <bits/stdc++.h>
using namespace std;

int totCount;
int firstCount;

// Function to get the total count
void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;

    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }

    firstCount = totCount;
    temp = num >> totCount;

    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}

// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
int flipBitsFromRightMostSetBit(int num)
{

    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);

    // XOR given number with the
    // number which is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);

    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);

    return num1;
}

// Driver Code
int main()
{
    int n = 120;

    cout << flipBitsFromRightMostSetBit(n)
         << endl;

    return 0;
}

// This code is contributed by divyeshrabadiya07

Java

// Java program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
import java.util.*;

class GFG{

static int totCount;
static int firstCount;

// Function to get the total count
static void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;

    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }

    firstCount = totCount;
    temp = num >> totCount;

    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}

// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
static int flipBitsFromRightMostSetBit(int num)
{

    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);

    // XOR given number with the
    // number which is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);

    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);

    return num1;
}

// Driver Code
public static void main (String[] args)
{
    int n = 120;

    System.out.println(
        flipBitsFromRightMostSetBit(n));
}
}

// This code is contributed by offbeat

Python3

# Python3 program to find the
# integer formed after flipping
# all bits to the left of the
# rightmost set bit

# Function to get the total count
def getTotCount(num):
    totCount = 1
    firstCount = 1

    temp = 1

    # Moving until we get
    # the rightmost set bit
    while (not(num & temp)):
        temp = temp << 1
        totCount += 1
    firstCount = totCount

    temp = num >> totCount

    # To get total number
    # of bits in a number
    while (temp):
        totCount += 1
        temp = temp >> 1

    return totCount, firstCount


# Function to find the integer formed
# after flipping all bits to the left
# of the rightmost set bit
def flipBitsFromRightMostSetBit(num):

    # Find the total count of bits and
    # the rightmost set bit
    totbit, firstbit = getTotCount(num)


    # XOR given number with the
    # number which is made up
    # of only totbits set

    num1 = num ^ ((1 << totbit) - 1)

    # To avoid flipping the bits
    # to the right of the set bit,
    # take XOR with the number
    # made up of only set firstbits

    num1 = num1 ^ ((1 << firstbit) - 1)

    return num1


if __name__=='__main__':
    n = 120
    print(flipBitsFromRightMostSetBit(n))

C#

// C# program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
using System;

class GFG{

static int totCount;
static int firstCount;

// Function to get the total count
static void getTotCount(int num)
{
    totCount = 1;
    firstCount = 1;
    int temp = 1;

    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }

    firstCount = totCount;
    temp = num >> totCount;

    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}

// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
static int flipBitsFromRightMostSetBit(int num)
{

    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);

    // XOR given number with the
    // number which is made up
    // of only totbits set
    int num1 = num ^ ((1 << totCount) - 1);

    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);

    return num1;
}

// Driver Code
public static void Main (string[] args)
{
    int n = 120;

    Console.Write(
        flipBitsFromRightMostSetBit(n));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// JavaScript program to find the
// integer formed after flipping
// all bits to the left of the
// rightmost set bit
let totCount;
let firstCount;

// Function to get the total count
function getTotCount(num)
{
    totCount = 1;
    firstCount = 1;
    let temp = 1;

    // Moving until we get
    // the rightmost set bit
    while ((num & temp) == 0)
    {
        temp = temp << 1;
        totCount += 1;
    }

    firstCount = totCount;
    temp = num >> totCount;

    // To get total number
    // of bits in a number
    while (temp != 0)
    {
        totCount += 1;
        temp = temp >> 1;
    }
}

// Function to find the integer formed
// after flipping all bits to the left
// of the rightmost set bit
function flipBitsFromRightMostSetBit(num)
{

    // Find the total count of bits and
    // the rightmost set bit
    getTotCount(num);

    // XOR given number with the
    // number which is made up
    // of only totbits set
    let num1 = num ^ ((1 << totCount) - 1);

    // To avoid flipping the bits
    // to the right of the set bit,
    // take XOR with the number
    // made up of only set firstbits
    num1 = num1 ^ ((1 << firstCount) - 1);

    return num1;
}

// Driver Code
let n = 120;

document.write(flipBitsFromRightMostSetBit(n));

// This code is contributed by sanjoy_62

</script>

Output:

8

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