LeetCode-in-Java
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 <= 52sconsists only of lowercase English letters.- Each letter appears in
sexactly twice. distance.length == 260 <= distance[i] <= 50
Solution
public class Solution {
public boolean checkDistances(String s, int[] distance) {
boolean[] seenFirstIndexYet = new boolean[26];
for (int idxIntoS = 0; idxIntoS < s.length(); ++idxIntoS) {
char c = s.charAt(idxIntoS);
if (!seenFirstIndexYet[c - 'a']) {
seenFirstIndexYet[c - 'a'] = true;
distance[c - 'a'] += idxIntoS;
} else {
// seenFirstIndexYet[c - 'a']
distance[c - 'a'] -= idxIntoS;
if (distance[c - 'a'] != -1) {
// early return
return false;
}
}
}
return true;
}
}