2399. Check Distances Between Same Letters

#### QUESTION:

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
``````

#### EXPLANATION:

easy的题目, 用dic去标记位置. 如果没有, 则表明是第一次出现, 那么就直接记录即可. 第二次出现的时候, 需要用第二次的位置减去第一次的位置, 就得到了距离. 最后再对距离和distance进行对比就可以得到结果.

#### SOLUTION:

``````class Solution {
func checkDistances(_ s: String, _ distance: [Int]) -> Bool {
var dic:Dictionary<Character,Int> = [:]
var arr:[Character] = Array(s)
for index in arr.indices {
var ch:Character = arr[index]
if dic[ch] == nil {
dic[ch] = index
} else {
dic[ch] = index - dic[ch]! - 1
}
}

for index in 0...25 {
var ch:Character = Character(UnicodeScalar(97 + index)!)
if (dic[ch] != nil) {
if (dic[ch] != distance[index]) {
return false
}
}
}
return true
}
}
``````