1012. Numbers With Repeated Digits

# Time:  O(logn)
# Space: O(logn)

class Solution(object):
    def numDupDigitsAtMostN(self, N):
        """
        :type N: int
        :rtype: int
        """
        def P(m, n):
            result = 1
            for _ in xrange(n):
                result *= m
                m -= 1
            return result

        digits = map(int, str(N+1))
        result = 0

        # Given 321
        #
        # 1. count numbers without repeated digits:
        # - X
        # - XX
        for i in xrange(1, len(digits)):
            result += P(9, 1)*P(9, i-1)
        
        # 2. count numbers without repeated digits:
        # - 1XX ~ 3XX
        # - 30X ~ 32X
        # - 320 ~ 321
        prefix_set = set()
        for i, x in enumerate(digits):
            for y in xrange(1 if i == 0 else 0, x):
                if y in prefix_set:
                    continue
                result += P(9-i, len(digits)-i-1)
            if x in prefix_set:
                break
            prefix_set.add(x)
        return N-result

Solution from kamyu104/LeetCode-Solutions · MIT