LeetCode 2262. Total Appeal of A String

原题链接在这里：https://leetcode.com/problems/total-appeal-of-a-string/

题目：

The appeal of a string is the number of distinct characters found in the string.

• For example, the appeal of "abbca" is 3 because it has 3 distinct characters: 'a', 'b', and 'c'.

Given a string s, return the total appeal of all of its substrings.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "abbca"
Output: 28
Explanation: The following are the substrings of "abbca":
- Substrings of length 1: "a", "b", "b", "c", "a" have an appeal of 1, 1, 1, 1, and 1 respectively. The sum is 5.
- Substrings of length 2: "ab", "bb", "bc", "ca" have an appeal of 2, 1, 2, and 2 respectively. The sum is 7.
- Substrings of length 3: "abb", "bbc", "bca" have an appeal of 2, 2, and 3 respectively. The sum is 7.
- Substrings of length 4: "abbc", "bbca" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 5: "abbca" has an appeal of 3. The sum is 3.
The total sum is 5 + 7 + 7 + 6 + 3 = 28.

Example 2:

Input: s = "code"
Output: 20
Explanation: The following are the substrings of "code":
- Substrings of length 1: "c", "o", "d", "e" have an appeal of 1, 1, 1, and 1 respectively. The sum is 4.
- Substrings of length 2: "co", "od", "de" have an appeal of 2, 2, and 2 respectively. The sum is 6.
- Substrings of length 3: "cod", "ode" have an appeal of 3 and 3 respectively. The sum is 6.
- Substrings of length 4: "code" has an appeal of 4. The sum is 4.
The total sum is 4 + 6 + 6 + 4 = 20.

Constraints:

• 1 <= s.length <= 105
• s consists of lowercase English letters.

题解：

For current char, how many points could it contribute. 

Total is the total points of appeal ending at current char.

There are 2 cases:

1. This char has never been appeared before, then it contributes to all the substring ending at this char.

2. This char has been appeard before, then it contributes to all the substring starting form its last appearence poisiton + 1 to this char.

Time Complexity: O(n). n = s.length().

Space: O(1).

AC Java:

class Solution {
    public long appealSum(String s) {
        int [] last = new int[26];
        long res = 0;
        int total = 0;
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            total += i + 1 - last[c - 'a'];
            last[c - 'a'] = i + 1;
            res += total;
        }
        
        return res;
    }
}

类似Count Unique Characters of All Substrings of a Given String.