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Python | Counting unset bits in a range

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Examples:

 Input: n = 42, l = 2, r = 5 Output: 2 (42) 10 = (1  0101  0) 2 There are ’2 ’unset bits in the range 2 to 5. Input: n = 80, l = 1, r = 4 Output: 4 

We have a solution to this problem, please refer to counter of unset bits in link bin (num) .

  • Now remove the first two characters of the output binary string, because the bin function by default prefixes "0b" to the output string.
  • Truncate the string starting at index ( l-1) to index r, and reverse it, then count the unset bits between them.
  • # Function for counting undefined bits in a range

      

    def unsetBits (n, l, r):

     

    # convert n to binary

     < / code> binary = bin (n)

     

    # remove first two characters

    binary = binary [ 2 :]

     

    # backward string

    binary = binary [ - 1 :: - 1 ]

     

     

    # read all unset bits "0" starting from index l-1

    # up to r, where r is exclusive

    print ( len ([binary [i] for i in range (l - 1 , r) if binary [i] = = ’0’ ]))

     
    # Driver program

    if __ name__ = = " __ main__ " :

      n = 42

    l = 2

    r = 5

    unsetBits (n, l, r)

    Output:

     2 

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