【leetcode】357. Count Numbers with Unique Digits【M】【72】

最新推荐文章于 2020-06-05 12:35:22 发布

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最新推荐文章于 2020-06-05 12:35:22 发布

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分类专栏： leetcode DP Python 算法文章标签： leetcode leetcode题解动态规划 python

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本文探讨如何计算在给定非负整数n的情况下，所有包含唯一数字的数的数量，其中0≤x<10^n。文章提供了一种使用回溯法的解决方案，并介绍了通过动态规划结合组合数学的方法来高效解决问题。

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Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n.

Example:
Given n = 2, return 91. (The answer should be the total numbers in the range of 0 ≤ x < 100, excluding [11,22,33,44,55,66,77,88,99])

Hint:

A direct way is to use the backtracking approach.
Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10n.
This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
Let f(k) = count of numbers with unique digits with length equals k.
f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.

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class Solution(object):
    def countNumbersWithUniqueDigits(self, n):
        if n == 0:
            return 1
            
        res = [10,81]
        
        for i in xrange(3,min(10,n)+1):
            j = 3
            t = 81
            
            while j <= i:
                t *= (9-j+2)
                j += 1
            res += t,
        #print res
        return sum(res[:min(n,10)])