LeetCode 2399. 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

• 我怎么就这么笨呢，，，光想着去处理第一个字符串，没想要可以结合对应的数组去处理，如果第2次出现，将之前的赋值为-1就不会出错了。。。。还是个简单的题目。

Runtime: 1 ms, faster than 99.30% of Java online submissions for Check Distances Between Same Letters.

Memory Usage: 42.4 MB, less than 61.34% of Java online submissions for Check Distances Between Same Letters.