0317MediumArrayHash TableBit Manipulation

Crates Handled by Both Crews

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Trains the technique from

LeetCode 2657Find the Prefix Common Array of Two Arrays

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.

Two crews unload the same n crates, tagged 1 through n. Each crew works through all the crates in its own order: crewA[i] is the crate the first crew handles in minute i, and crewB[i] is the crate the second crew handles in that same minute. Both lists are permutations of 1 through n.

At the end of every minute the yard manager writes down how many crates have by then been handled by both crews, counting everything either crew touched from minute 0 up to and including the current minute.

Return the manager's log as a list of n numbers, where entry i is the figure recorded at the end of minute i.

Examples

Example 1

Input: crewA = [4, 1, 2, 3], crewB = [1, 4, 3, 2]
Output: [0, 2, 2, 4]

After minute 0 the crews have handled crates {4} and {1}, with nothing in common. After minute 1 they hold {4, 1} and {1, 4}, so both crates count. Minute 2 adds crate 2 for the first crew and crate 3 for the second, neither of which the other crew has yet. Minute 3 finishes every crate.

Example 2

Input: crewA = [2, 1, 5, 3, 4], crewB = [2, 5, 1, 4, 3]
Output: [1, 1, 3, 3, 5]

Both crews start on crate 2, so the first figure is 1. Minute 1 adds crate 1 and crate 5, still leaving only crate 2 shared. Minute 2 gives the first crew crate 5 and the second crew crate 1, so crates 1, 2 and 5 are now shared.

Example 3

Input: crewA = [3, 2, 1, 4], crewB = [4, 1, 2, 3]
Output: [0, 0, 2, 4]

Nothing overlaps for the first two minutes. After minute 2 the crews hold {3, 2, 1} and {4, 1, 2}, sharing crates 1 and 2.

Example 4

Input: crewA = [1], crewB = [1]
Output: [1]

With one crate, both crews handle it in minute 0.

Constraints

1 <= crewA.length == crewB.length == n <= 50
1 <= crewA[i], crewB[i] <= n
crewA and crewB are both permutations of the integers 1 through n.

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 shared_progress(crewA: list[int], crewB: list[int]) -> list[int]:

Java

public int[] sharedProgress(int[] crewA, int[] crewB)

The intended approach

Time O(n)Space O(n) for the tallies and the returned log

Keep one tally per crate tag counting how many crews have touched it so far. Each minute, bump the tally for the crate the first crew handled and for the crate the second crew handled; a tally that reaches 2 means that crate has just become shared, so a running total goes up by one. When both crews pick the same crate in the same minute the tally climbs from 0 to 2 through 1, which still triggers exactly one increment. Record the running total after each minute.

Pythonentry point: shared_progress

def shared_progress(crewA, crewB):
    # O(n) time: constant work per minute. O(n) space for tallies plus the log.
    n = len(crewA)
    tally = [0] * (n + 1)
    shared = 0
    log = []
    for i in range(n):
        tally[crewA[i]] += 1
        if tally[crewA[i]] == 2:
            shared += 1
        tally[crewB[i]] += 1
        if tally[crewB[i]] == 2:
            shared += 1
        log.append(shared)
    return log

The obvious version

For every minute, rebuild both prefixes as sets from scratch and take the size of their intersection.

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_shared_progress(crewA, crewB):
    out = []
    for i in range(len(crewA)):
        left = set(crewA[:i + 1])
        right = set(crewB[:i + 1])
        out.append(len(left & right))
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
4 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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