821. Shortest Distance to a Character

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

String (Chuỗi)

⏱️ Thời gian O(n)

💾 Không gian O(n)

💻 Lời Giải

C++ 0821-shortest-distance-to-a-character.cpp

class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        queue<int> mq;
        const int n = s.size();
        vector<int> dist(n, 0);
        for (int i = 0; i < n; ++i) {
            if (s[i] == c) {
                mq.push(i);
            }
        }
        while (!mq.empty()) {
            int m = mq.size();
            while (m--) {
                int i = mq.front();
                mq.pop();
                int r = i + 1;
                int l = i - 1;
                if (0 <= l and l < n and s[l] != c and dist[l] == 0) {
                    dist[l] = 1 + dist[i];
                    mq.push(l);
                }
                if (0 <= r and r < n and s[r] != c and dist[r] == 0) {
                    dist[r] = 1 + dist[i];
                    mq.push(r);
                }
            }
        }
        return dist;
    }
};