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Java-based LeetCode algorithm problem solutions, regularly updated
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2399\. Check Distances Between Same Letters
Easy
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`
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