855. Count Unique Characters Of All Substrings Of A Given String¶

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

Hard

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.

Solution¶

class Solution {
    private HashMap<Character, TreeSet<Integer>> map;
    public int uniqueLetterString(String s) {
        int n = s.length();
        map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            char current = s.charAt(i);
            if (!map.containsKey(current))
                map.put(current, new TreeSet<>());
            map.get(current).add(i);
        }

        int ans = 0;
        for (int i = 0; i < n; i++) {
            char current = s.charAt(i);
            TreeSet<Integer> currentSet = map.get(current);
            Integer leftIdx = currentSet.lower(i);
            Integer rightIdx = currentSet.higher(i);


            int leftCount = 0, rightCount = 0, totalCount = 0;
            if (leftIdx != null) 
                leftCount = i - leftIdx - 1;
            else 
                leftCount += i;
            if (rightIdx != null) 
                rightCount += rightIdx - i - 1;
            else 
                rightCount += n - i - 1;

            ans += (leftCount * (rightCount + 1));
            ans += (rightCount);
            ans++;
        }
        return ans;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here