1127MediumArrayHash TableLinked List

Dropping the Banned Tags From the Chain

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Trains the technique from

LeetCode 3217Delete Nodes From Linked List Present in Array

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 run of codes banned is given, no two of them the same, and a chain of tags follows.

Because the harness passes plain JSON, the chain reaches you as chain, listing the tag codes in order from the front.

Drop every tag whose code turns up in banned and return what is left, again as a plain list read from the front. At least one tag always survives.

Examples

Example 1

Input: banned = [1, 3, 5], chain = [1, 2, 3, 4, 5, 6]
Output: [2, 4, 6]

The three odd codes are banned, so the three even tags survive in the order they stood.

Example 2

Input: banned = [7], chain = [7, 1, 7, 2, 7, 3]
Output: [1, 2, 3]

Every tag coded 7 goes, whichever position it held, and the others keep their order.

Example 3

Input: banned = [6], chain = [1, 2, 3, 4, 5]
Output: [1, 2, 3, 4, 5]

No tag carries the banned code, so the chain comes back untouched.

Constraints

1 <= banned.length <= 10^5
1 <= banned[i] <= 10^5
no two codes in banned are the same
1 <= chain.length <= 10^5
1 <= chain[i] <= 10^5
at least one tag carries a code missing from banned

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 modified_list(banned: list[int], chain: list[int]) -> list[int]:

Java

public int[] modifiedList(int[] banned, int[] chain)

The intended approach

Time O(n + m) over the two lists.Space O(m) for the set of banned codes.

Asking whether a tag's code is banned is the whole job, and doing it by walking the banned run for every tag costs the two lengths multiplied. Gathering the banned codes into a set first turns each question into a single lookup, so the work drops to one pass over each list. After that the walk is straightforward: read the chain from the front and keep the tags whose codes are absent from the set, in the order they came. Every tag is judged by its own code alone, so a code appearing several times in the chain has every one of its tags dropped, not merely the first.

Pythonentry point: modified_list

def modified_list(banned, chain):
    blocked = set(banned)
    out = []
    for code in chain:
        if code not in blocked:
            out.append(code)
    return out

The obvious version

For each tag in turn, walk the whole banned run looking for its code and keep the tag only when the walk finds nothing. Costs the chain length times the banned length.

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_modified_list(banned, chain):
    out = []
    for code in chain:
        found = False
        for other in banned:
            if other == code:
                found = True
                break
        if not found:
            out.append(code)
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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