Count Unique Char Substrs

// https://leetcode.com/problems/count-unique-characters-of-all-substrings-of-a-given-string/description/

/*
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.

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

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

Input: s = "LEETCODE"
Output: 92

*/

/*
Intuition
Let's think about how a character can be found as a unique character.

Think about string "XAXAXXAX" and focus on making the second "A" a unique character.
We can take "XA(XAXX)AX" and between "()" is our substring.
We can see here, to make the second "A" counted as a uniq character, we need to:

insert "(" somewhere between the first and second A
insert ")" somewhere between the second and third A
For step 1 we have "A(XA" and "AX(A", 2 possibility.
For step 2 we have "A)XXA", "AX)XA" and "AXX)A", 3 possibilities.

So there are in total 2 * 3 = 6 ways to make the second A a unique character in a substring.
In other words, there are only 6 substring, in which this A contribute 1 point as unique string.

Instead of counting all unique characters and struggling with all possible substrings,
we can count for every char in S, how many ways to be found as a unique char.
We count and sum, and it will be out answer.


Explanation
index[26][2] record last two occurrence index for every upper characters.
Initialise all values in index to -1.
Loop on string S, for every character c, update its last two occurrence index to index[c].
Count when loop. For example, if "A" appears twice at index 3, 6, 9 seperately, we need to count:
For the first "A": (6-3) * (3-(-1))"
For the second "A": (9-6) * (6-3)"
For the third "A": (N-9) * (9-6)"
*/

#include <vector>
#include <string>

using namespace std;

// Time: O(n), Space: O(1)
int uniqueLetterString(string S) {
    int res = 0, n = S.size(), M = 1e9 + 7;
    vector<vector<int>> idx(26, vector<int>(2, -1));
    for (int i = 0; i < n; ++i) {
        int c = S[i] - 'A';
        res = (res + (i - idx[c][1]) * (idx[c][1] - idx[c][0]) % M) % M;
        idx[c][0] = idx[c][1];
        idx[c][1] = i;
    }
    for (int c = 0; c < 26; ++c) {
        res = (res + (n - idx[c][1]) * (idx[c][1] - idx[c][0]) % M) % M;
    }
    return res;
}```