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

• Difficulty Level : Medium
• Last Updated : 01 Oct, 2021

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

Input: N = 120
Output:
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 ``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 has 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 has 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 has 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 has 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

 ``

Output:

`8`

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

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