0089EasyTwo PointersStringString MatchingZ AlgorithmKnuth–Morris–Pratt AlgorithmBoyer–Moore String-Search Algorithm

Locate the Routing Tag

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Trains the technique from

LeetCode 28Find the Index of the First Occurrence 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.

Every crate in a depot is stencilled with a long identifier, given as the lowercase string label. A conveyor belt decides where a crate goes by hunting for a short routing marker inside that identifier, given as the lowercase string tag.

Return the smallest index i such that the len(tag) consecutive characters of label starting at position i spell tag exactly. If the marker is nowhere inside the identifier, return -1.

Both strings are non-empty, and the marker may well be longer than the identifier.

Examples

Example 1

Input: label = "abracadabra", tag = "cad"
Output: 4

Characters 4 through 6 of the identifier spell the marker; nothing earlier does.

Example 2

Input: label = "mississippi", tag = "issip"
Output: 4

A shorter partial match begins at position 1 and dies at the fourth character, so the answer is the later position 4 where all five characters line up.

Example 3

Input: label = "banana", tag = "nab"
Output: -1

The three characters never sit next to each other in that arrangement.

Constraints

1 <= label.length <= 10^4
1 <= tag.length <= 10^4
label and tag contain only 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 locate_tag(label: str, tag: str) -> int:

Java

public int locateTag(String label, String tag)

The intended approach

Time O(n + m)Space O(m)

Precompute, for every prefix of `tag`, the length of its longest proper prefix that is also a suffix of it. Scanning `label` once, a mismatch then falls back along that table instead of rewinding into `label`, so each character of the identifier is examined a constant number of times.

Pythonentry point: locate_tag

def locate_tag(label, tag):
    n = len(label)
    m = len(tag)
    if m > n:
        return -1
    fallback = [0] * m
    k = 0
    for i in range(1, m):
        while k > 0 and tag[i] != tag[k]:
            k = fallback[k - 1]
        if tag[i] == tag[k]:
            k += 1
        fallback[i] = k
    k = 0
    for i in range(n):
        ch = label[i]
        while k > 0 and ch != tag[k]:
            k = fallback[k - 1]
        if ch == tag[k]:
            k += 1
            if k == m:
                return i - m + 1
    return -1

The obvious version

Line the marker up against every start position and compare character by character, O(n * 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_locate_tag(label, tag):
    n = len(label)
    m = len(tag)
    for i in range(n - m + 1):
        hit = True
        for j in range(m):
            if label[i + j] != tag[j]:
                hit = False
                break
        if hit:
            return i
    return -1

How this problem was checked

Reference solution executed against 10 test cases — 10 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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