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,
- Convert decimal into binary using bin(num) function.
- Now remove first two characters of output binary string because bin function appends ‘0b’ as prefix in output string by default.
- Slice string starting from index (l-1) to index r and reverse it, then count unset bits in between.
The set bits in the binary form of a number (obtained using the bin() method) can be obtained using the count() method.
Time Complexity: O(1)
Auxiliary Space: O(1)