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:
Constraints:
2 <= s.length <= 52s consists only of lowercase English letters.s exactly twice.distance.length == 260 <= distance[i] <= 50| Constructor and Description |
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Solution() |
public boolean checkDistances(String s, int[] distance)
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