2399

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

Related Topics:
Array, Hash Table, String

Similar Questions:

• Two Sum (Easy)
• Shortest Distance to a Character (Easy)

Solution 1. Prefix State Map

// OJ: https://leetcode.com/problems/check-distances-between-same-letters
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(26)
class Solution {
public:
    bool checkDistances(string s, vector<int>& A) {
        int index[26] = {[0 ... 25] = -1};
        for (int i = 0; i < s.size(); ++i) {
            int c = s[i] - 'a';
            if (index[c] == -1) index[c] = i;
            else if (A[c] != i - index[c] - 1) return false;
        }
        return true;
    }
};