0844MediumArrayHash TableStringSorting

Patching a Label in One Pass

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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 label reads s. A batch of patches is described by three lists of the same length: patch i targets position indices[i] of the label, looks for the text sources[i] starting exactly there, and if it finds it, replaces that text with targets[i].

A patch whose sources[i] does not appear at that exact position is discarded. Every patch is judged against the original label, never against the result of another patch, and the targeted stretches of two patches never overlap.

Return the label after every surviving patch is applied.

Examples

Example 1

Input: s = "abcdef", indices = [0, 4], sources = ["ab", "ef"], targets = ["wx", "yz"]
Output: "wxcdyz"

Both patches find their text where they expect it, so "ab" at position 0 becomes "wx" and "ef" at position 4 becomes "yz", leaving "cd" untouched.

Example 2

Input: s = "abcdef", indices = [0, 4], sources = ["ab", "gg"], targets = ["wx", "yz"]
Output: "wxcdef"

Position 4 holds "ef", not "gg", so that patch is discarded and only the first one is applied.

Example 3

Input: s = "abab", indices = [0, 2], sources = ["ab", "ba"], targets = ["q", "r"]
Output: "qab"

Position 0 holds "ab" so it becomes "q". Position 2 holds "ab", not "ba", so that patch is discarded.

Constraints

1 <= s.length <= 1000
1 <= indices.length <= 100
indices.length == sources.length
indices.length == targets.length
0 <= indices[i] <= 999
1 <= sources[i].length <= 50
1 <= targets[i].length <= 50
s consists of lowercase English letters only
sources[i] and targets[i] consist of lowercase English letters only
Every index is inside s, and the stretches two patches target never overlap

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 patch_label(s: str, indices: list[int], sources: list[str], targets: list[str]) -> str:

Java

public String patchLabel(String s, int[] indices, String[] sources, String[] targets)

The intended approach

Time O(L + sum of source lengths) for a label of length L.Space O(L) for the recorded patches and the rebuilt label.

Every patch is judged against the original label, so the checks are independent and can all be made up front. Record, for each position that a surviving patch starts at, how much text it consumes and what it writes. Then walk the label once: at a recorded position write the replacement and skip past the consumed text, and otherwise copy one character across. Because the targeted stretches never overlap, one pass is enough and no position is ever rewritten twice.

Pythonentry point: patch_label

def patch_label(s, indices, sources, targets):
    plan = {}
    for at, want, give in zip(indices, sources, targets):
        if s[at:at + len(want)] == want:
            plan[at] = (len(want), give)
    out = []
    i = 0
    n = len(s)
    while i < n:
        if i in plan:
            eaten, give = plan[i]
            out.append(give)
            i += eaten
        else:
            out.append(s[i])
            i += 1
    return "".join(out)

The obvious version

Work out which patches survive, then rebuild the label by walking every position and, at each, searching the whole surviving patch list for one that starts there. Re-scans the patch list for every position instead of indexing them once.

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_patch_label(s, indices, sources, targets):
    keep = []
    for k in range(len(indices)):
        at = indices[k]
        want = sources[k]
        if s[at:at + len(want)] == want:
            keep.append((at, want, targets[k]))
    out = ""
    i = 0
    while i < len(s):
        hit = None
        for at, want, give in keep:
            if at == i:
                hit = (want, give)
        if hit is None:
            out += s[i]
            i += 1
        else:
            out += hit[1]
            i += len(hit[0])
    return out

How this problem was checked

Reference solution executed against 13 test cases — 13 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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