0104MediumHash TableStringSliding Window

Mosaic Bundle Positions

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Trains the technique from

LeetCode 438Find All Anagrams in a String

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 tiler is repairing a mosaic border. The finished border is written down as border, one lowercase letter per tile in the order the tiles run left to right, each letter standing for a colour. A sealed box of spare tiles is written down the same way as bundle.

The tiler wants to know where a single box could have supplied a run of the border. A box supplies the run starting at position i when that run is exactly as long as the box and holds the same colours in the same quantities, in whatever order they happen to be laid. Quantities matter: a box holding two red tiles cannot supply a run holding one red tile and two blue ones.

Return the starting positions of every run the box could supply, listed from smallest position to largest. Return an empty list when there is no such run, which includes the case of a box holding more tiles than the whole border.

Examples

Example 1

Input: border = "tsrrtsr", bundle = "rst"
Output: [0, 3, 4]

The runs beginning at 0, 3 and 4 each hold one t, one s and one r. The run beginning at 4 reaches the end of the border, so the last run counts as well.

Example 2

Input: border = "rssrrs", bundle = "rrs"
Output: [2, 3]

The box holds two r tiles and one s. The run beginning at 0 draws on the same two colours but in the wrong quantities, so only the runs at 2 and 3 qualify.

Example 3

Input: border = "mm", bundle = "mmm"
Output: []

The box holds three tiles and the border only has two, so no run can be as long as the box.

Constraints

1 <= border.length <= 3 * 10^4
1 <= bundle.length <= 3 * 10^4
border and bundle consist of lowercase English letters
bundle may hold more tiles than border, in which case the answer is empty

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 bundle_positions(border: str, bundle: str) -> list[int]:

Java

public List<Integer> bundlePositions(String border, String bundle)

The intended approach

Time O(n)Space O(1)

Only 26 colours exist, so hold one counter per colour recording how many tiles of that colour the box still owes the current run, plus a tally of how many colours are currently off balance. Slide a run of the box's length along the border, crediting the tile that enters and refunding the tile that leaves, adjusting the off-balance tally only when a counter crosses zero. A run qualifies exactly when nothing is off balance.

Pythonentry point: bundle_positions

def bundle_positions(border, bundle):
    m = len(bundle)
    n = len(border)
    if m > n:
        return []
    base = ord("a")
    deficit = [0] * 26
    for ch in bundle:
        deficit[ord(ch) - base] += 1
    wrong = 0
    for count in deficit:
        if count != 0:
            wrong += 1
    spots = []
    for i in range(n):
        k = ord(border[i]) - base
        if deficit[k] == 0:
            wrong += 1
        deficit[k] -= 1
        if deficit[k] == 0:
            wrong -= 1
        if i >= m:
            j = ord(border[i - m]) - base
            if deficit[j] == 0:
                wrong += 1
            deficit[j] += 1
            if deficit[j] == 0:
                wrong -= 1
        if i >= m - 1 and wrong == 0:
            spots.append(i - m + 1)
    return spots

The obvious version

Sort the box once, then sort every run of that length and compare, O(n * m log m).

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_bundle_positions(border, bundle):
    m = len(bundle)
    target = sorted(bundle)
    spots = []
    for i in range(len(border) - m + 1):
        if sorted(border[i:i + m]) == target:
            spots.append(i)
    return spots

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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