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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).