Python | Count unset bits in a range

Given a non-negative number n and two values l and r. The problem is to count the number of unset bits in the range l to r in the binary representation of n, i.e, to count unset bits from the rightmost lth bit to the rightmost rth bit. Examples:

Input : n = 42, l = 2, r = 5
Output : 2
(42)10 = (101010)2
There are '2' unset bits in the range 2 to 5.

Input : n = 80, l = 1, r = 4
Output : 4

We have existing solution for this problem please refer Count unset bits in a range link. We can solve this problem quickly in Python. Approach is very simple,

1. Convert decimal into binary using bin(num) function.
2. Now remove first two characters of output binary string because bin function appends ‘0b’ as prefix in output string by default.
3. Slice string starting from index (l-1) to index r and reverse it, then count unset bits in between.

Python3

# Function to count unset bits in a range

def unsetBits(n,l,r):

    # convert n into it's binary
    binary = bin(n)

    # remove first two characters
    binary = binary[2:]

    # reverse string
    binary = binary[-1::-1]

    # count all unset bit '0' starting from index l-1
    # 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

Another Approach:

The set bits in the binary form of a number (obtained using the bin() method) can be obtained using the count() method.

Python3

# Function to count set bits in a range

def countUnsetBits(n, l, r):

    # convert n into its binary form
        # using bin()
    # and then process it using string
    # slice methods
    binary = bin(n)[-1:1:-1]

    # count all set bit '1' starting from index l-1
    # to r, where r is exclusive
    print(binary[l - 1: r].count("0"))


# Driver Code
n = 42
l = 2
r = 5
countUnsetBits(n, l, r)


#This code is contributed by phasing17

Output

2

Time Complexity: O(1)

Auxiliary Space: O(1)