0250MediumArrayHash TableStringTrie

Longest Shared Route Lead

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Trains the technique from

LeetCode 3043Find the Length of the Longest Common Prefix

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 courier network stamps every parcel with a numeric routing code, written in decimal with no leading zeros. A code is read from the left: the first digit names the region, the digit after it names a depot inside that region, and so on down to the delivery walk. Two codes therefore travel together for as far as their leading digits agree.

You are given two arrays of routing codes, northern and southern. Take one code from northern and one from southern. Their shared lead is how many digits they agree on, counting from the leftmost digit and stopping at the first position where they differ or where either code runs out of digits.

Return the longest shared lead over every way of pairing one code from northern with one code from southern. If no such pairing agrees on even its leftmost digit, return 0.

Examples

Example 1

Input: northern = [917, 91, 4], southern = [9174, 63]
Output: 3

Pairing 917 with 9174 agrees on the digits 9, 1 and 7, then 9174 still has a digit left while 917 has run out, so that pairing has a shared lead of 3.

Example 2

Input: northern = [52, 61], southern = [7, 8]
Output: 0

Every pairing differs at the leftmost digit, so no pairing has a shared lead at all.

Example 3

Input: northern = [8, 80], southern = [8888, 800]
Output: 2

Pairing 80 with 800 agrees on the digits 8 and 0, and 80 then runs out, giving a shared lead of 2.

Constraints

1 <= northern.length, southern.length <= 5 * 10^4
1 <= northern[i], southern[i] <= 10^8
Codes are written in decimal without leading zeros

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 longest_shared_lead(northern: list[int], southern: list[int]) -> int:

Java

public int longestSharedLead(int[] northern, int[] southern)

The intended approach

Time O((m + n) * d^2) where m and n are the array lengths and d <= 9 is the digit width, since each of the d leads of a code is a string of up to d characters to build and hashSpace O(m * d^2) for the stored leads

Write each northern code out as digits and drop every one of its leading digit runs into a hash set, so the set holds each length of lead any northern code offers. Then walk each southern code's digits from the left, extending its lead one digit at a time while the set still contains it and stopping as soon as it does not, since a longer lead cannot survive where a shorter one already failed. Track the greatest length reached.

Pythonentry point: longest_shared_lead

def longest_shared_lead(northern, southern):
    leads = set()
    for code in northern:
        text = str(code)
        for end in range(1, len(text) + 1):
            leads.add(text[:end])
    best = 0
    for code in southern:
        text = str(code)
        for end in range(1, len(text) + 1):
            if text[:end] not in leads:
                break
            if end > best:
                best = end
    return best

The obvious version

Compare every northern code against every southern code digit by digit and keep the largest agreement, O(m * n * d).

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_longest_shared_lead(northern, southern):
    best = 0
    for first in northern:
        left = str(first)
        for second in southern:
            right = str(second)
            shared = 0
            while (shared < len(left) and shared < len(right)
                   and left[shared] == right[shared]):
                shared += 1
            if shared > best:
                best = shared
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 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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