# Check Distances Between Same Letters

Easy

1

0.0% Acceptance

In this lab, you need to implement a function named `checkDistances`

that receives 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**.

Your task is to return `true`

if `s`

is a **well-spaced** string, otherwise return `false`

.

**Example 1:**

`const s = "abaccb"; const 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]; checkDistances(s, distance); // 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:**

`const s = "aa"; const 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]; checkDistances(s, distance); // 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`