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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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