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.
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
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. Let it be count.
- Calculate ans = (r – l + 1) – count.
- Return ans.
- Python | Count unset bits in a range
- Unset 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
- Count unset bits of a number
- Count set bits in a range
- Python | Count set bits in a range
- Range query for count of set bits
- Count number of set bits in a range using bitset
- Unset the last m bits
- Find the largest number with n set and m unset bits
- Find the smallest number with n set and m unset bits
- Check if bits of a number has count of consecutive set bits in increasing order
- Check if a number has same number of set and unset bits
- Toggle bits in the given range
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Improved By : Mithun Kumar