1923. Longest Common Subpath

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].

Similar Questions:

• Reconstruct Itinerary (Medium)

Solution 1. Rolling Hash + Binary Answer

// OJ: https://leetcode.com/problems/longest-common-subpath/
// Author: github.com/lzl124631x
// Time: O(N + logM * P) where P is the total length of all paths.
// Space: O(M)
class Solution {
    bool valid(vector<vector<int>> &A, int len) {
        unordered_map<unsigned long long, int> m;
        unordered_set<unsigned long long> s; // Note that we must use `unsigned long long` to reduce the chance of hash conflict
        unsigned d = 16777619;
        for (auto &v : A) {
            s.clear();
            unsigned long long h = 0, p = 1;
            for (int i = 0; i < v.size(); ++i) {
                h = h * d + v[i];
                if (i >= len) h -= v[i - len] * p;
                else p *= d;
                if (i >= len - 1) s.insert(h);
            }
            for (auto x : s) {
                if (++m[x] == A.size()) return true;
            }
        }
        return false;
    }
public:
    int longestCommonSubpath(int n, vector<vector<int>>& A) {
        int minLen = INT_MAX;
        for (auto &v : A) minLen = min(minLen, (int)v.size());
        int L = 0, R = minLen;
        while (L < R) {
            int M = (L + R + 1) / 2;
            if (valid(A, M)) L = M;
            else R = M - 1;
        }
        return L;
    }
};

Or

// OJ: https://leetcode.com/problems/longest-common-subpath/
// Author: github.com/lzl124631x
// Time: O(N + logM * P) where P is the total length of all paths.
// Space: O(M)
class Solution {
    bool valid(vector<vector<int>> &A, int len) {
        unordered_set<unsigned long long> s, tmp;
        for (int i = 0; i < A.size() && (i == 0 || s.size()); ++i) {
            unsigned long long d = 1099511628211, h = 0, p = 1;
            tmp.clear();
            swap(s, tmp);
            for (int j = 0; j < A[i].size(); ++j) {
                h = h * d + A[i][j];
                if (j < len) p *= d;
                else h -= A[i][j - len] * p;
                if (j >= len - 1 && (i == 0 || tmp.count(h))) s.insert(h);
            }
        }
        return s.size();
    }
public:
    int longestCommonSubpath(int n, vector<vector<int>>& A) {
        int L = 0, R = min_element(begin(A), end(A), [](auto &a, auto &b) { return a.size() < b.size(); })->size();
        while (L < R) {
            int M = (L + R + 1) / 2;
            if (valid(A, M)) L = M;
            else R = M - 1;
        }
        return L;
    }
};