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:

Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 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))

chevron_right


Output:

8


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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.