Is Count the Number of Special Characters II easy or hard?

The problem is classified as Medium difficulty. The main challenge is recognizing the ordering constraint between lowercase and uppercase occurrences and storing positions efficiently so the validation step can be done in constant time per character.

Count the Number of Special Characters II Python/Java solution

Python and Java implementations commonly use dictionaries or arrays to record the first uppercase and last lowercase positions. After scanning the string once, a loop over the alphabet counts valid characters. Both implementations achieve O(n) time and O(1) space complexity.

How to solve Count the Number of Special Characters II in O(n)?

Scan the string and store two values for each letter: the last index of its lowercase form and the first index of its uppercase form. After the scan, iterate over all 26 characters and count those where both exist and the lowercase index is smaller than the uppercase index. This ensures all lowercase occurrences appear before the first uppercase occurrence.

What is the best approach for Count the Number of Special Characters II?

The best approach runs in O(n) time by recording the last index of each lowercase letter and the first index of each uppercase letter. After scanning the string once, check whether the last lowercase index occurs before the first uppercase index for each character. This guarantees the required ordering while keeping space usage constant.

Is Count the Number of Special Characters II asked at Google/Amazon/Meta?

Problems involving character tracking, case comparisons, and hash table lookups frequently appear in interviews at companies like Amazon, Google, and Meta. This problem tests string traversal, indexing logic, and efficient use of constant-sized data structures, which are common interview patterns.

What data structure is used in Count the Number of Special Characters II?

The solution typically uses a hash table or fixed-size arrays to store character positions. Hash maps provide flexible key lookups, while arrays of size 26 offer faster indexing because the alphabet size is constant.

What is the time complexity of Count the Number of Special Characters II?

The optimal solution runs in O(n) time where n is the length of the string. Each character is processed once to record its position, and then a constant loop over 26 letters verifies the conditions. Space complexity remains O(1) because only fixed-size arrays or maps for the alphabet are used.

Description Hints 1 Solution Video 14:27

3121. Count the Number of Special Characters II

Medium60.3% acceptance3 companies

You are given a string word. A letter c is called special if it appears both in lowercase and uppercase in word, and every lowercase occurrence of c appears before the first uppercase occurrence of c.

Return the number of special letters in word.

Example 1:

Input: word = "aaAbcBC"

Output: 3

Explanation:

The special characters are 'a', 'b', and 'c'.

Example 2:

Input: word = "abc"

Output: 0

Explanation:

There are no special characters in word.

Example 3:

Input: word = "AbBCab"

Output: 0

Explanation:

There are no special characters in word.

Constraints:

1 <= word.length <= 2 * 10⁵
word consists of only lowercase and uppercase English letters.

Asked at 3 companies

Google Amazon Microsoft

Topics

Hash Table String

Hints open one at a time

Hint 1

For each character c, store the first occurrence of its uppercase and the last occurrence of its lowercase.

14:27

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3121. Count the Number of Special Characters II

Asked at 3 companies

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