Introduction

In this blog post, we will explore how to solve the LeetCode problem "1923" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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].

Explanation

Here's a solution approach, complexity analysis, and Python code for finding the longest common subpath:

• Binary Search for Length: Use binary search to find the length of the longest common subpath. The search space is from 0 to the minimum length of any path.
• Rolling Hash for Subpath Comparison: For each potential length, use rolling hash to efficiently compute the hash values of all subpaths of that length for each friend's path.

• Set Intersection: Check if there is a common subpath (hash value) among all friends by taking the intersection of the sets of hash values for each friend.

• Runtime Complexity: O(N log N), where N is the sum of the lengths of all paths.

• Storage Complexity: O(N)

Code

    def longestCommonSubpath(n: int, paths: list[list[int]]) -> int:
    """
    Finds the length of the longest common subpath shared by all paths.

    Args:
        n: The number of cities.
        paths: A list of paths, where each path is a list of city indices.

    Returns:
        The length of the longest common subpath, or 0 if no common subpath exists.
    """

    def check(length: int) -> bool:
        """
        Checks if there is a common subpath of the given length.
        """
        if length == 0:
            return True

        hash_sets = []
        for path in paths:
            if len(path) < length:
                return False

            p = 10**9 + 7  # A large prime number
            x = 256  # A constant (can be any number > n)
            x_pow = 1
            for _ in range(length - 1):
                x_pow = (x_pow * x) % p

            path_hash_set = set()
            hash_val = 0
            for i in range(length):
                hash_val = (hash_val * x + path[i]) % p

            path_hash_set.add(hash_val)

            for i in range(length, len(path)):
                hash_val = (hash_val - path[i - length] * x_pow) % p
                hash_val = (hash_val * x + path[i]) % p
                if hash_val < 0:
                    hash_val += p  # Ensure hash_val is non-negative
                path_hash_set.add(hash_val)
            hash_sets.append(path_hash_set)

        common_hashes = hash_sets[0]
        for i in range(1, len(hash_sets)):
            common_hashes = common_hashes.intersection(hash_sets[i])

        return len(common_hashes) > 0

    min_len = min(len(path) for path in paths)
    left, right = 0, min_len
    ans = 0

    while left <= right:
        mid = (left + right) // 2
        if check(mid):
            ans = mid
            left = mid + 1
        else:
            right = mid - 1

    return ans