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

• 我怎么就這么笨呢，，，光想著去處理第一個(gè)字符串，沒想要可以結(jié)合對應(yīng)的數(shù)組去處理，如果第2次出現(xiàn)，將之前的賦值為-1就不會出錯(cuò)了。。。。還是個(gè)簡單的題目。

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.