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
Approach: Following are the steps:
- Calculate num = ((1 << r) – 1) ^ ((1 << (l-1)) – 1). This will produce a number num having r number of bits and bits in the range l to r are the only set bits.
- Count number of set bits in the number (n & num). Refer this post.
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- Python | Count set bits in a range
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- Python | Count unset bits in a range
- Check if bits of a number has count of consecutive set bits in increasing order
- Set bits in N equals to M in the given range.
- Check whether all the bits are set in the given range
- Unset bits in the given range
- Copy set bits in a range
- Set all the bits in given range of a number
- Toggle bits in the given range
- Check whether all the bits are unset in the given range
- Check whether all the bits are unset in the given range or not
- Check whether bits are in alternate pattern in the given range
Improved By : jit_t