CheckDistances-LCQ.cpp

/*
Check Distances Between Same Letters

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
*/

class Solution {
public:
	bool checkDistances(string s, vector<int>& distance) {
		map<char, int>mp;
		for (int i = 0; i < s.size(); i++) {
			if (mp.find(s[i]) != mp.end()) {
				int val = mp[s[i]];
				mp[s[i]] = abs(val - i) - 1;
			} else {
				mp[s[i]] = i;
			}
		}
		for (int i = 0; i < s.size(); i++) {
			int tmp = s[i] - 'a';
			if (mp[s[i]] != distance[tmp]) {
				return false;
			}
		}
		return true;
	}
};