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Formatted question description: https://leetcode.ca/all/2262.html

2262. Total Appeal of A String

  • Difficulty: Hard.
  • Related Topics: Hash Table, String, Dynamic Programming.
  • Similar Questions: Count Unique Characters of All Substrings of a Given String, Count Vowel Substrings of a String, Vowels of All Substrings.

Problem

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.

Solution

  • class Solution {
    public long appealSum(String s) {
        int len = s.length();
        int[] lastPos = new int[26];
        Arrays.fill(lastPos, -1);
        long res = 0;
        for (int i = 0; i < len; i++) {
            int idx = s.charAt(i) - 'a';
            res += (i - lastPos[idx]) * (len - i);
            lastPos[idx] = i;
        }
        return res;
    }
}

  • Todo

  • class Solution:
    def appealSum(self, s: str) -> int:
        ans = t = 0
        pos = [-1] * 26
        for i, c in enumerate(s):
            c = ord(c) - ord('a')
            t += i - pos[c]
            ans += t
            pos[c] = i
        return ans

############

# 2262. Total Appeal of A String
# https://leetcode.com/problems/total-appeal-of-a-string/

class Solution:
    def appealSum(self, s: str) -> int:
        n = len(s)
        
        def go(k):
            count = total = 0
            
            for x in s:
                if x == k:
                    total += (count * (count + 1)) // 2
                    count = 0
                else:
                    count += 1
            
            total += (count * (count + 1)) // 2
            
            return (n * (n + 1)) // 2 - total
        
        res = 0
        for k in string.ascii_lowercase:
            res += go(k)
            
        return res

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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