0067MediumArrayHash TableSliding Window

Two-Sack Pickup Run

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Trains the technique from

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 row of parcel lockers is numbered from 0. Locker i holds one parcel whose destination is written as the route code codes[i].

A courier works the row with two sacks. A sack is unlabelled until the first parcel drops into it; that parcel fixes the sack's route code, and from then on the sack accepts only parcels carrying the same code. Neither sack has a capacity limit.

The courier picks any locker as the starting point, empties it, steps to the next locker on the right, empties that one, and keeps going without skipping lockers. The run stops when the lockers run out, or when the parcel in the next locker carries a code that matches neither sack and both sacks are already labelled.

Return the largest number of parcels the courier can carry away, taken over every possible starting locker.

Examples

Example 1

Input: codes = [4, 4, 7, 7, 7, 1, 1, 3]
Output: 5

Starting at locker 0 collects codes 4, 4, 7, 7, 7 before code 1 blocks the run, giving 5 parcels. Starting at locker 2 also reaches 5, and no start does better.

Example 2

Input: codes = [6, 6, 1, 6, 1, 1, 2, 2]
Output: 6

Starting at locker 0 the sacks take code 6 and code 1, which covers lockers 0 through 5 for 6 parcels; code 2 at locker 6 then ends the run.

Example 3

Input: codes = [3, 1, 4, 1, 5]
Output: 3

The best run is lockers 1 through 3 with codes 1, 4, 1, so 3 parcels.

Constraints

1 <= codes.length <= 10^5
0 <= codes[i] < codes.length

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 longest_pickup(codes: list[int]) -> int:

Java

public int longestPickup(int[] codes)

The intended approach

Time O(n)Space O(1)

A run is a contiguous block of lockers holding at most two different route codes, so the task is the longest such block. Slide a window across the row, keeping a tally of how many lockers inside it carry each code; whenever the tally grows to three codes, advance the left edge and decrement its code until one code drops to zero and leaves the tally. Every position of the right edge yields the longest legal run ending there, and the best of those is the answer.

Pythonentry point: longest_pickup

def longest_pickup(codes):
    tally = {}
    left = 0
    best = 0
    for right in range(len(codes)):
        code = codes[right]
        tally[code] = tally.get(code, 0) + 1
        while len(tally) > 2:
            leaving = codes[left]
            tally[leaving] -= 1
            if tally[leaving] == 0:
                del tally[leaving]
            left += 1
        if right - left + 1 > best:
            best = right - left + 1
    return best

The obvious version

Try every starting locker and walk rightwards collecting the distinct codes seen, stopping that start as soon as a third code shows up.

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_longest_pickup(codes):
    n = len(codes)
    best = 0
    for start in range(n):
        kinds = set()
        for end in range(start, n):
            kinds.add(codes[end])
            if len(kinds) > 2:
                break
            if end - start + 1 > best:
                best = end - start + 1
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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