0771MediumHash TableString

Sections Drafted Before Being Approved

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Trains the technique from

LeetCode 3121Count the Number of Special Characters II

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 document has up to 26 sections, each tagged by one letter of the alphabet. An audit trail arrives as the string log, one character per touch, read left to right in the order the touches happened:

a lowercase letter is a draft touch on the section with that tag;
the same letter uppercase is an approval touch on that same section.

A section is settled when the trail shows at least one draft touch on it, at least one approval touch on it, and no draft touch on it after its earliest approval touch.

Return how many sections are settled.

Examples

Example 1

Input: log = "ppPqrQR"
Output: 3

Section p is drafted at positions 0 and 1 and approved at position 2. Section q is drafted at position 3 and approved at position 5, and section r is drafted at position 4 and approved at position 6. All three have every draft touch ahead of their first approval.

Example 2

Input: log = "mMm"
Output: 0

Section m is drafted at position 0, approved at position 1, then drafted again at position 2. That final draft touch comes after the approval, so the section is not settled.

Example 3

Input: log = "ffFgGf"
Output: 1

Section g is drafted at position 3 and approved at position 4, so it is settled. Section f is drafted at positions 0, 1 and 5 and approved at position 2, and the touch at position 5 falls after that approval.

Constraints

1 <= log.length <= 2 * 10^5
log consists only of lowercase and uppercase English letters.

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 count_promoted_letters(log: str) -> int:

Java

public int countPromotedLetters(String log)

The intended approach

Time O(n) for a trail of n touches.Space O(1) beyond the input, since at most 26 letters are tracked.

A section is settled exactly when its final draft touch sits earlier in the trail than its earliest approval touch, so only two positions per letter matter. One pass over the trail keeps overwriting the last draft position for each lowercase letter and records the first approval position for each uppercase letter, then a walk over the drafted letters counts the ones whose two recorded positions are in that order.

Pythonentry point: count_promoted_letters

def count_promoted_letters(log):
    # One pass recording, per section letter, the position of its last draft touch
    # and of its first approval touch. Time O(n), space O(1) since at most 26
    # letters are ever tracked.
    last_draft = {}
    first_signoff = {}
    for i in range(len(log)):
        ch = log[i]
        if "a" <= ch <= "z":
            last_draft[ch] = i
        else:
            key = chr(ord(ch) + 32)
            if key not in first_signoff:
                first_signoff[key] = i
    total = 0
    for ch, pos in last_draft.items():
        if ch in first_signoff and pos < first_signoff[ch]:
            total += 1
    return total

The obvious version

Take the 26 letters one at a time. Collect every position of its draft touches and every position of its approval touches, then require both lists to be non-empty and check every draft position against every approval position directly. Quadratic in the trail length.

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_count_promoted_letters(log):
    total = 0
    for k in range(26):
        low = chr(97 + k)
        up = chr(65 + k)
        drafts = []
        signoffs = []
        for i in range(len(log)):
            if log[i] == low:
                drafts.append(i)
            elif log[i] == up:
                signoffs.append(i)
        if not drafts or not signoffs:
            continue
        clean = True
        for d in drafts:
            for s in signoffs:
                if d > s:
                    clean = False
                    break
            if not clean:
                break
        if clean:
            total += 1
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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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