Easy Bit Manipulation No Iterations from m to n

  • 0
    class Solution(object):
    def rangeBitwiseAnd(self, m, n):
        count = 0
        while (m != n):
            m >>= 1; n >>= 1; count += 1
        return n << count

    We only need to find the common prefix of m and n. Keep a count of how many bits we shift and shift the prefix leftwards that many times.

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