longest-common-subpath

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1923. Longest Common Subpath (Hard)

There is a country of n cities numbered from 0 to n - 1. In this country, there is a road connecting every pair of cities.

There are m friends numbered from 0 to m - 1 who are traveling through the country. Each one of them will take a path consisting of some cities. Each path is represented by an integer array that contains the visited cities in order. The path may contain a city more than once, but the same city will not be listed consecutively.

Given an integer n and a 2D integer array paths where paths[i] is an integer array representing the path of the ith friend, return the length of the longest common subpath that is shared by every friend's path, or 0 if there is no common subpath at all.

A subpath of a path is a contiguous sequence of cities within that path.

 

Example 1:

Input: n = 5, paths = [[0,1,2,3,4],
                       [2,3,4],
                       [4,0,1,2,3]]
Output: 2
Explanation: The longest common subpath is [2,3].

Example 2:

Input: n = 3, paths = [[0],[1],[2]]
Output: 0
Explanation: There is no common subpath shared by the three paths.

Example 3:

Input: n = 5, paths = [[0,1,2,3,4],
                       [4,3,2,1,0]]
Output: 1
Explanation: The possible longest common subpaths are [0], [1], [2], [3], and [4]. All have a length of 1.

 

Constraints:

• 1 <= n <= 105
• m == paths.length
• 2 <= m <= 105
• sum(paths[i].length) <= 105
• 0 <= paths[i][j] < n
• The same city is not listed multiple times consecutively in paths[i].

Related Topics

[Array] [Binary Search] [Suffix Array] [Hash Function] [Rolling Hash]

Hints

Hint 1

If there is a common path with length x, there is for sure a common path of length y where y < x.

Hint 2

We can use binary search over the answer with the range [0, min(path[i].length)].

Hint 3

Using binary search, we want to verify if we have a common path of length m. We can achieve this using hashing.