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2262. Total Appeal of A String

Description

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.

 

Solutions

Solution: Dynamic Programming

  • Time complexity: O(n)
  • Space complexity: O(26 -> 1)

 

JavaScript

js
/**
 * @param {string} s
 * @return {number}
 */
const appealSum = function (s) {
  const n = s.length;
  const BASE_CODE = 'a'.charCodeAt(0);
  const lastSeen = Array.from({ length: 26 }, () => -1);
  let dp = 0;
  let result = 0;

  for (let index = 0; index < n; index++) {
    const code = s[index].charCodeAt(0) - BASE_CODE;

    dp += index - lastSeen[code];
    result += dp;
    lastSeen[code] = index;
  }

  return result;
};

Released under the MIT license