0910EasyHash TableString

Letters Stamped Both Ways

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Trains the technique from

LeetCode 3120Count the Number of Special Characters I

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A plate reads word, a mix of lowercase and uppercase English letters.

A letter of the alphabet is stamped both ways when the plate holds it in lowercase somewhere and in uppercase somewhere. Return how many letters of the alphabet are stamped both ways.

Examples

Example 1

Input: word = "kEttLeKnObs"
Output: 2

The letter k appears as both k and K, and the letter e appears as both e and E. The letters l, o, t, n, b and s appear in only one case each.

Example 2

Input: word = "mMnNoOpPqQ"
Output: 5

All five letters turn up in both cases.

Example 3

Input: word = "abcdefghij"
Output: 0

Nothing on the plate is uppercase, so no letter is stamped both ways.

Constraints

1 <= word.length <= 50
word consists of lowercase and uppercase English letters only

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def number_of_special_chars(word: str) -> int:

Java

public int numberOfSpecialChars(String word)

The intended approach

Time O(n) for a plate of length n, one pass.Space O(1): two records of twenty-six flags.

Where the two cases sit on the plate does not matter, only that each turns up at all, so one pass recording which lowercase letters were seen and which uppercase letters were seen is enough. The answer is then how many letters appear in both records, which is the size of their overlap.

Pythonentry point: number_of_special_chars

def number_of_special_chars(word):
    lower = set()
    upper = set()
    for ch in word:
        if "a" <= ch <= "z":
            lower.add(ch)
        else:
            upper.add(ch.lower())
    return len(lower & upper)

The obvious version

Take each of the twenty-six letters in turn and scan the whole plate twice, once looking for its lowercase form and once for its uppercase, counting the letter when both turn up. Reads the plate fifty-two times over and keeps nothing between letters.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_number_of_special_chars(word):
    total = 0
    for code in range(26):
        low = chr(97 + code)
        up = chr(65 + code)
        seen_low = False
        for ch in word:
            if ch == low:
                seen_low = True
        seen_up = False
        for ch in word:
            if ch == up:
                seen_up = True
        if seen_low and seen_up:
            total += 1
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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