0077HardHash TableStringSliding Window

Shortest Kit Stretch

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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 conveyor carries parts past an assembly cell in a fixed order. The string belt records the stamp on each part, one letter per part, reading along the belt. A kit order is written as the string kit: every letter in it is a stamp the kit needs, and a letter written twice means the kit needs two parts carrying that stamp. Stamps are case sensitive, so a part stamped b cannot fill a slot that asks for B.

Find the shortest unbroken stretch of the belt that can fill the whole kit and return that stretch as a string. Extra parts inside the stretch are set aside and cost nothing, but a stretch may not skip over parts. If several stretches tie for shortest, return the one that starts earliest along the belt. If no stretch can fill the kit, return an empty string.

The cell is fed at line rate, so your work must stay proportional to the combined length of belt and kit.

Examples

Example 1

Input: belt = "PQRSTQP", kit = "PQ"
Output: "PQ"

The first two parts already carry both stamps the kit asks for, and no single part can fill a two-stamp kit.

Example 2

Input: belt = "aabbcc", kit = "abcb"
Output: "abbc"

The kit lists b twice, so the stretch has to reach across both b parts as well as one a and one c.

Example 3

Input: belt = "abxba", kit = "ab"
Output: "ab"

Two stretches of length two can fill the kit, and the earlier of the two is the answer.

Example 4

Input: belt = "xy", kit = "z"
Output: ""

No part on the belt carries the stamp z, so nothing can fill the kit.

Constraints

m == belt.length
n == kit.length
1 <= m, n <= 10^5
belt and kit consist of uppercase and lowercase 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 shortest_kit_stretch(belt: str, kit: str) -> str:

Java

public String shortestKitStretch(String belt, String kit)

The intended approach

Time O(m + n)Space O(1), at most 52 stamp counters

Keep a window over the belt together with a per-stamp tally of how many copies it still owes the kit, plus a single counter of unfilled slots. Extend the right edge until nothing is owed, then pull the left edge in as far as the window stays fillable, recording the window whenever it is strictly shorter than the best so far, which keeps the earliest of any tied stretches. Each edge only ever moves right, so both edges cross the belt once.

Pythonentry point: shortest_kit_stretch

def shortest_kit_stretch(belt, kit):
    want = {}
    for ch in kit:
        want[ch] = want.get(ch, 0) + 1
    owed = len(kit)
    best_len = len(belt) + 1
    best_start = 0
    left = 0
    for right in range(len(belt)):
        ch = belt[right]
        if ch in want:
            if want[ch] > 0:
                owed -= 1
            want[ch] -= 1
        while owed == 0:
            if right - left + 1 < best_len:
                best_len = right - left + 1
                best_start = left
            dropped = belt[left]
            if dropped in want:
                want[dropped] += 1
                if want[dropped] > 0:
                    owed += 1
            left += 1
    if best_len > len(belt):
        return ""
    return belt[best_start:best_start + best_len]

The obvious version

Try stretch lengths in increasing order and, for each length, every start position from the left, tallying that stretch from scratch. The first stretch that fills the kit is by construction the shortest, earliest one.

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_shortest_kit_stretch(belt, kit):
    want = {}
    for ch in kit:
        want[ch] = want.get(ch, 0) + 1
    n = len(belt)
    for length in range(1, n + 1):
        for start in range(0, n - length + 1):
            have = {}
            for ch in belt[start:start + length]:
                have[ch] = have.get(ch, 0) + 1
            ok = True
            for ch, needed in want.items():
                if have.get(ch, 0) < needed:
                    ok = False
                    break
            if ok:
                return belt[start:start + length]
    return ""

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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