leetcode-2399-easy

Check Distances Between Same Letters

You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.

Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25).

In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.

Return true if s is a well-spaced string, otherwise return false.

Example 1:

Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.
Example 2:

Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.
Constraints:

2 <= s.length <= 52
s consists only of lowercase English letters.
Each letter appears in s exactly twice.
distance.length == 26
0 <= distance[i] <= 50

思路一：遍历字符，记录每个字符的位置，然后对比给定的数组是否匹配

public boolean checkDistances(String s, int[] distance) {
    Map<Character, Integer> map = new HashMap<>();

    for (int i = 0; i < s.length(); i++) {
        int finalI = i;
        map.compute(s.charAt(i), (k, v) -> v == null ? finalI : finalI - v - 1);
    }

    for (Map.Entry<Character, Integer> e : map.entrySet()) {
        Character c = e.getKey();
        if (distance[c - 'a'] != e.getValue()) return false;
    }

    return true;
}

思路二：优化思路一，在遍历字符时就可以提前判断字符位置是否正确

public boolean checkDistances(String s, int[] distance) {
    Map<Character, Integer> map = new HashMap<>();

    for (int i = 0; i < s.length(); i++) {
        char key = s.charAt(i);
        if (map.containsKey(key)) {
            if (i - map.get(key) != distance[key - 'a']) return false;
        } else {
            map.put(key, i);
        }
    }

    return true;
}