Let's define a function countUniqueChars(s)
that returns the number of unique characters on s
.
 For example, calling
countUniqueChars(s)
ifs = "LEETCODE"
then"L"
,"T"
,"C"
,"O"
,"D"
are the unique characters since they appear only once ins
, thereforecountUniqueChars(s) = 5
. Given a strings
, return the sum ofcountUniqueChars(t)
wheret
is a substring ofs
. The test cases are generated such that the answer fits in a 32bit 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 <= 10^{5}
s
consists of uppercase English letters only.
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.

