Package g2301_2400.s2399_check_distances_between_same_letters

Class Solution

java.lang.Object
g2301_2400.s2399_check_distances_between_same_letters.Solution
public class Solution extends Object
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

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • checkDistances

      public boolean checkDistances(String s, int[] distance)