0102MediumHash TableTwo PointersStringSliding Window

Seed Row Packet Window

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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 seed catalogue codes every variety with a single lowercase letter. The string packet lists the seeds inside one packet, and the string row lists, in planting order, the seeds a gardener has already put in the ground along a long bed.

The gardener wants to know whether some unbroken stretch of the bed was planted from exactly one packet's worth of seed: a stretch qualifies when it holds the same varieties as packet in the same quantities, planted in whatever order. Return true if such a stretch exists in row, and false if none does.

Quantities count. A packet with two a seeds does not match a stretch holding one a and two b seeds, even though both draw on the same two varieties.

Examples

Example 1

Input: packet = "abc", row = "zzbca"
Output: true

The three seeds planted at positions 2 through 4 are one of each variety, matching the packet in a different order.

Example 2

Input: packet = "aab", row = "zabbz"
Output: false

Every three-seed stretch either includes a z or holds one a and two b seeds, while the packet holds two a and one b.

Example 3

Input: packet = "ba", row = "xxab"
Output: true

The qualifying stretch is the final pair of seeds, so the last stretch of the bed has to be examined too.

Constraints

1 <= packet.length <= 10^4
1 <= row.length <= 10^4
packet and row consist of lowercase English letters
packet may be longer than row, in which case no stretch can qualify

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 contains_packet_window(packet: str, row: str) -> bool:

Java

public boolean containsPacketWindow(String packet, String row)

The intended approach

Time O(n)Space O(1)

A qualifying stretch has exactly the packet's length, so slide a window of that fixed width along the bed. Keep a tally of the 26 varieties inside the window, adding the seed that enters and subtracting the one that leaves, and compare the tally with the packet's tally once the window is full width. Each seed is added and removed once, and the tallies are a fixed 26 counters wide.

Pythonentry point: contains_packet_window

def contains_packet_window(packet, row):
    width = len(packet)
    if width > len(row):
        return False
    wanted = [0] * 26
    for seed in packet:
        wanted[ord(seed) - 97] += 1
    window = [0] * 26
    for i, seed in enumerate(row):
        window[ord(seed) - 97] += 1
        if i >= width:
            window[ord(row[i - width]) - 97] -= 1
        if i >= width - 1 and window == wanted:
            return True
    return False

The obvious version

Sort the packet once, then sort every stretch of that width in the bed and compare. O(n * w log w).

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_contains_packet_window(packet, row):
    width = len(packet)
    wanted = sorted(packet)
    for start in range(len(row) - width + 1):
        if sorted(row[start:start + width]) == wanted:
            return True
    return False

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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