0646MediumHash TableStringSliding Window

Bay Coverage in the Parcel Log

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Trains the technique from

LeetCode 1358Number of Substrings Containing All Three Characters

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 parcel hub sends every parcel down one of three chutes, which the crew label r, g and b. The shift log is a string log where the character at each position names the chute used by that parcel, in the order the parcels went through.

A stretch is any run of one or more consecutive positions of log. A stretch is covering when at least one parcel inside it went down each of the three chutes.

Count the covering stretches. Two stretches are different when they begin at different positions or end at different positions, even when they read the same. Return that count.

Examples

Example 1

Input: log = "gbrgb"
Output: 6

Numbering the positions from 1, the covering stretches are 1-3, 1-4, 1-5, 2-4, 2-5 and 3-5. Each of those six holds a parcel for chute r, one for chute g and one for chute b.

Example 2

Input: log = "rrggrr"
Output: 0

Chute b never appears in the log, so no stretch can hold a parcel for all three chutes.

Example 3

Input: log = "brrrbg"
Output: 4

Chute g is used only by the last parcel, so a covering stretch has to end at position 6. The four that also reach back to an r and a b are 1-6, 2-6, 3-6 and 4-6.

Constraints

3 <= log.length <= 5 * 10^4
log consists only of the characters 'r', 'g' and 'b'.

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_full_stretches(log: str) -> int:

Java

public int countFullStretches(String log)

The intended approach

Time O(n) where n is the length of log; each position is visited once and the three stored positions are compared in constant time.Space O(1): three positions and a running total.

Sweep the log once from left to right, keeping the most recent position at which each of the three chutes was used. Once all three have been seen, the earliest of those three positions decides how many stretches ending at the current position are covering: every start index from the beginning of the log up to and including that earliest position works, and none after it does. Adding that count at each step totals the answer in one pass.

Pythonentry point: count_full_stretches

def count_full_stretches(log):
    # last[c] holds the most recent position at which bay c was recorded, -1 if
    # bay c has not been seen yet. O(n) time, O(1) space.
    last = {"r": -1, "g": -1, "b": -1}
    total = 0
    for i in range(len(log)):
        last[log[i]] = i
        earliest = min(last["r"], last["g"], last["b"])
        if earliest >= 0:
            total += earliest + 1
    return total

The obvious version

Fix every start position, then walk right one parcel at a time keeping a tally of how many parcels went down each chute so far. Whenever the tally holds all three chutes, add one to the count.

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_full_stretches(log):
    n = len(log)
    total = 0
    for i in range(n):
        tally = {}
        for j in range(i, n):
            tally[log[j]] = tally.get(log[j], 0) + 1
            if len(tally) == 3:
                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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