Shortest Distance to a Character — LeetCode 821 Python Solution

The problem

Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s. The distance between two indices i and j is abs(i - j), where abs is the absolute value function.

Example

Input

s = "loveleetcode", c = "e"

Output

[3,2,1,0,1,0,0,1,2,2,1,0]

Explanation

The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).

Python solution

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        n = len(s)
        ans = [n] * n
        pre = -inf
        for i, ch in enumerate(s):
            if ch == c:
                pre = i
            ans[i] = min(ans[i], i - pre)
        suf = inf
        for i in range(n - 1, -1, -1):
            if s[i] == c:
                suf = i
            ans[i] = min(ans[i], suf - i)
        return ans

Complexity

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 821. Shortest Distance to a Character is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Two Pointers.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

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