Given a non-negative number n and two values l and r. The problem is to count the number of set bits in the range l to r in the binary representation of n, i.e, to count set bits from the rightmost lth bit to the rightmost rth bit.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Input : n = 42, l = 2, r = 5 Output : 2 (42)10 = (101010)2 There are '2' set bits in the range 2 to 5. Input : n = 79, l = 1, r = 4 Output : 4
We have existing solution for this problem please refer Count set 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 set bits in between.
- Python | Count unset bits in a range
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- Count number of set bits in a range using bitset
- Python Bin | Count total bits in a number
- Count set bits using Python List comprehension
- Python map function | Count total set bits in all numbers from 1 to n
- Check if bits of a number has count of consecutive set bits in increasing order
- Unset bits in the given range
- Check whether all the bits are set in the given range
- Set all the bits in given range of a number
- Toggle bits in the given range
- Copy set bits in a range
- Set bits in N equals to M in the given range.
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