LeetCode 0828. Count Unique Characters of All Substrings of a Given String Solution in Java, C++, Python & More | Explanation + Code

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.

 

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) × (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.

PythonJavaC++GoTypeScriptRust

class Solution: def uniqueLetterString(self, s: str) -> int: d = defaultdict(list) for i, c in enumerate(s): d[c].append(i) ans = 0 for v in d.values(): v = [-1] + v + [len(s)] for i in range(1, len(v) - 1): ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]) return ans(code-box)

class Solution { public int uniqueLetterString(String s) { List<Integer>[] d = new List[26]; Arrays.setAll(d, k -> new ArrayList<>()); for (int i = 0; i < 26; ++i) { d[i].add(-1); } for (int i = 0; i < s.length(); ++i) { d[s.charAt(i) - 'A'].add(i); } int ans = 0; for (var v : d) { v.add(s.length()); for (int i = 1; i < v.size() - 1; ++i) { ans += (v.get(i) - v.get(i - 1)) * (v.get(i + 1) - v.get(i)); } } return ans; } }(code-box)

class Solution { public: int uniqueLetterString(string s) { vector<vector<int>> d(26, {-1}); for (int i = 0; i < s.size(); ++i) { d[s[i] - 'A'].push_back(i); } int ans = 0; for (auto& v : d) { v.push_back(s.size()); for (int i = 1; i < v.size() - 1; ++i) { ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]); } } return ans; } };(code-box)

func uniqueLetterString(s string) (ans int) { d := make([][]int, 26) for i := range d { d[i] = []int{-1} } for i, c := range s { d[c-'A'] = append(d[c-'A'], i) } for _, v := range d { v = append(v, len(s)) for i := 1; i < len(v)-1; i++ { ans += (v[i] - v[i-1]) * (v[i+1] - v[i]) } } return }(code-box)

function uniqueLetterString(s: string): number { const d: number[][] = Array.from({ length: 26 }, () => [-1]); for (let i = 0; i < s.length; ++i) { d[s.charCodeAt(i) - 'A'.charCodeAt(0)].push(i); } let ans = 0; for (const v of d) { v.push(s.length); for (let i = 1; i < v.length - 1; ++i) { ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]); } } return ans; }(code-box)

impl Solution { pub fn unique_letter_string(s: String) -> i32 { let mut d: Vec<Vec<i32>> = vec![vec![-1; 1]; 26]; for (i, c) in s.chars().enumerate() { d[(c as usize) - ('A' as usize)].push(i as i32); } let mut ans = 0; for v in d.iter_mut() { v.push(s.len() as i32); for i in 1..v.len() - 1 { ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]); } } ans as i32 } }(code-box)