Number of Digit One — LeetCode 233 Python Solution

The problem

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example

Input

n = 13

Output

6

Python solution

class Solution:
    def countDigitOne(self, n: int) -> int:
        @cache
        def dfs(i: int, cnt: int, limit: bool) -> int:
            if i >= len(s):
                return cnt
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                ans += dfs(i + 1, cnt + (j == 1), limit and j == up)
            return ans

        s = str(n)
        return dfs(0, 0, True)

Complexity

Pattern: Dynamic Programming

Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 233. Number of Digit One is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.

The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.

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