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821. Shortest Distance to a Character

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.

 

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

Solution 1

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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

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class Solution {
    public int[] shortestToChar(String s, char c) {
        int n = s.length();
        int[] ans = new int[n];
        final int inf = 1 << 30;
        Arrays.fill(ans, inf);
        for (int i = 0, pre = -inf; i < n; ++i) {
            if (s.charAt(i) == c) {
                pre = i;
            }
            ans[i] = Math.min(ans[i], i - pre);
        }
        for (int i = n - 1, suf = inf; i >= 0; --i) {
            if (s.charAt(i) == c) {
                suf = i;
            }
            ans[i] = Math.min(ans[i], suf - i);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        int n = s.size();
        const int inf = 1 << 30;
        vector<int> ans(n, inf);
        for (int i = 0, pre = -inf; i < n; ++i) {
            if (s[i] == c) {
                pre = i;
            }
            ans[i] = min(ans[i], i - pre);
        }
        for (int i = n - 1, suf = inf; ~i; --i) {
            if (s[i] == c) {
                suf = i;
            }
            ans[i] = min(ans[i], suf - i);
        }
        return ans;
    }
};

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func shortestToChar(s string, c byte) []int {
    n := len(s)
    ans := make([]int, n)
    const inf int = 1 << 30
    pre := -inf
    for i := range s {
        if<