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841. Shortest Distance To A Character

Array Two Pointers String

Problem - Shortest Distance To A Character

Easy

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 1:

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).
The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.

Example 2:

Input: s = "aaab", c = "b"
Output: [3,2,1,0]

 

Constraints:

  • 1 <= s.length <= 104
  • s[i] and c are lowercase English letters.
  • It is guaranteed that c occurs at least once in s.

Solutions

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class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        n = len(s)
        result = [n] * n
        last = float('-inf')
        next_val = float('inf')

        for i, char in enumerate(s):
            if char == c:
                last = i
            result[i] = min(result[i], i - last)

        for i in range(n - 1, -1, -1):
            if s[i] == c:
                next_val = i
            result[i] = min(result[i], next_val - i)

        return result

Submission Stats:

  • Runtime: 1 ms (75.50%)
  • Memory: 18 MB (25.64%)