821. Shortest Distance to a Character

1. Description

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.

2. Example

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]

3. Constraints

• 1 <= s.length <= $10^4$
• s[i] and c are lowercase English letters.
• It is guaranteed that c occurs at least once in s.

4. Solutions

Two Pointers

n = str.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    vector<int> shortestToChar(const string &str, char c) {
        vector<int> result(str.size());
        for (int i = 0, left = 0, right = 0; i < str.size(); ++i) {
            for (; i <= right && right < str.size() && str[right] != c; ++right)
                ;

            result[i] =
                min(str[left] == c ? i - left : INT_MAX,
                    right < str.size() && str[right] == c ? right - i : INT_MAX);

            if (i == right) {
                left = right;
                ++right;
            }
        }

        return result;
    }
};