LeetCode 821. Shortest Distance to a Character

Description

https://leetcode.com/problems/shortest-distance-to-a-character/

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) = 3.
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.

Explanation

Find c indexes, and find the minimum distance between each character to c indexes.

Python Solution

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        
        c_indexes = []
        
        for i, s_c in enumerate(s):
            if s_c == c:
                c_indexes.append(i)
                
        results = []    
        for i, s_c in enumerate(s):
            min_distance = sys.maxsize
            for c_i in c_indexes:
                min_distance = min(abs(i - c_i), min_distance)
                                
            results.append(min_distance)
                   
        return results
        
        
  • Time Complexity: O(N^2)
  • Space Complexity: O(N)

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