0420MediumStringStackSimulation

Cut the Marker from the Tape

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Trains the technique from

LeetCode 1910Remove All Occurrences of a Substring

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 label machine has printed a long paper tape carrying the lowercase letters of s, one letter per cell, and the operator has to strip a faulty marker part out of it.

The operator repeats one step for as long as the marker still appears anywhere on the tape:

Find the leftmost run of cells whose letters spell part, that is the match beginning at the smallest cell index.
Cut those cells out and splice the two loose ends of the tape together.

Splicing can bring letters together that were far apart before, and the marker may well turn up across a fresh splice, in which case it is cut too. The operator stops only when no run of cells anywhere on the tape spells part.

Return the letters left on the tape when the operator stops, as a string. The result may be the empty string.

Examples

Example 1

Input: s = "aabb", part = "ab"
Output: ""

The leftmost match sits at cells 1 and 2. Cutting it splices the remaining `a` and `b` together, and that pair spells the marker as well, so it goes too and nothing is left.

Example 2

Input: s = "pqrpqqrpqr", part = "pqr"
Output: "pqqr"

The leftmost match is at the start. Cutting it leaves `pqqrpqr`, whose leftmost match is the final three cells, and cutting that leaves `pqqr`, which holds no match.

Example 3

Input: s = "abc", part = "d"
Output: "abc"

No cell spells the marker, so the operator makes no cut and the tape is unchanged.

Example 4

Input: s = "ababa", part = "aba"
Output: "ba"

Two runs spell the marker, one starting at cell 0 and one at cell 2, and they share cells. The operator takes the one starting at cell 0, leaving `ba`, which holds no match.

Example 5

Input: s = "aaa", part = "aa"
Output: "a"

The leftmost match covers cells 0 and 1, and cutting it leaves a single `a`.

Constraints

1 <= s.length <= 1000
1 <= part.length <= 1000
s and part consist of lowercase English letters only.

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 remove_occurrences(s: str, part: str) -> str:

Java

public String removeOccurrences(String s, String part)

The intended approach

Time O(n * m) for a tape of length n and a marker of length m: every cell is wound on once and each winding compares at most m letters.Space O(n) for the spool, which is also the returned tape.

Wind the tape onto a spool one cell at a time. Each time a cell is added, compare the last `part.length` cells on the spool against the marker and unwind exactly that many when they agree. Because a cut is only ever tested at the spool's end, the letters brought together by the cut sit at the end too and the very next cell added re-tests them, so splices formed by earlier cuts are handled without any rescanning.

Pythonentry point: remove_occurrences

def remove_occurrences(s, part):
    # Feed the tape through one character at a time onto a growing spool. After
    # every character check whether the spool now ends with the marker, and if
    # so unwind exactly those characters; the splice this creates is examined by
    # the next character, so no cut is ever missed.
    width = len(part)
    spool = []
    for letter in s:
        spool.append(letter)
        if len(spool) >= width and spool[len(spool) - width:] == list(part):
            del spool[len(spool) - width:]
    return "".join(spool)

The obvious version

Do exactly what the operator does: search the whole tape for the first match, rebuild the tape without those cells, and search again from scratch, until no match is found.

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_remove_occurrences(s, part):
    tape = s
    at = tape.find(part)
    while at != -1:
        tape = tape[:at] + tape[at + len(part):]
        at = tape.find(part)
    return tape

How this problem was checked

Reference solution executed against 16 test cases — 16 passed
5 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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