2262 - Total Appeal of A String (Hard)

Problem Link

https://leetcode.com/problems/total-appeal-of-a-string/

Problem Statement

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 <= 10^5
• s consists of lowercase English letters.

Approach 1: Hash Map

Observation: The appeal of all substrings ending at $i$-th is that ending at $i - 1$th plus the number of substrings that don't contain $s[i]$, which can be calculated by tracking the last occurrence.

Written by @wingkwong

class Solution {
public:
    long long appealSum(string s) {
        long long ans = 0, cnt = 1, n = s.size();
        unordered_map<int, int> last;
        last[s[0]] = 1;
        for (int i = 0; i < n; i++) {
            cnt += i - last[s[i]] + 1;
            last[s[i]] = i + 1;
            ans += cnt;
        }
        return ans;
    }
};