828. 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)
ifs = "LEETCODE"
then"L"
,"T"
,"C"
,"O"
,"D"
are the unique characters since they appear only once ins
, thereforecountUniqueChars(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
Solution 1: Calculate the Contribution of Each Character
For each character $c_i$ in the string $s$, when it appears only once in a substring, it contributes to the count of unique characters in that substring.
Therefore, we only need to calculate for each character $c_i$, how many substrings contain this character only once.
We use a hash table or an array $d$ of length $26$, to store the positions of each character in $s$ in order of index.
For each character $c_i$, we iterate through each position $p$ in $d[c_i]$, find the adjacent positions $l$ on the left and $r$ on the right, then the number of substrings that meet the requirements by expanding from position $p$ to both sides is $(p - l) \times (r - p)$. We perform this operation for each character, add up the contributions of all characters, and get the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.
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