README.md

828. Count Unique Characters of All Substrings of a Given String

Let's define a function countUniqueChars(s) that returns the number of unique characters in s.

• For example, calling countUniqueChars(s) if s = "LEETCODE" then "L", "T", "C", "O", "D" are the unique characters since they appear only once in s, therefore countUniqueChars(s) = 5.

Given a string s, return the sum of countUniqueChars(t) where t is a substring of s. The test cases are generated such that the answer fits in a 32-bit integer.

Notice that some substrings can be repeated so in this case you have to count the repeated ones too.

Example 1:

Input: s = "ABC"
Output: 10
Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".
Every substring is composed with only unique letters.
Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10

Example 2:

Input: s = "ABA"
Output: 8
Explanation: The same as example 1, except countUniqueChars("ABA") = 1.

Example 3:

Input: s = "LEETCODE"
Output: 92

Constraints:

• 1 <= s.length <= 105
• s consists of uppercase English letters only.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn unique_letter_string(s: String) -> i32 {
        let s = s.as_bytes();
        let mut indices = vec![vec![-1]; 26];
        let mut ret = 0;

        for i in 0..s.len() {
            indices[(s[i] - b'A') as usize].push(i as i32);
        }

        for i in 0..26 {
            indices[i].push(s.len() as i32);

            for j in 1..indices[i].len() - 1 {
                ret += (indices[i][j] - indices[i][j - 1]) * (indices[i][j + 1] - indices[i][j]);
            }
        }

        ret
    }
}