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1923. Longest Common Subpath

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Description

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].

Solutions

Python3

Java

class Solution {
    int N = 100010;
    long[] h = new long[N];
    long[] p = new long[N];
    private int[][] paths;
    Map<Long, Integer> cnt = new HashMap<>();
    Map<Long, Integer> inner = new HashMap<>();

    public int longestCommonSubpath(int n, int[][] paths) {
        int left = 0, right = N;
        for (int[] path : paths) {
            right = Math.min(right, path.length);
        }
        this.paths = paths;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (check(mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private boolean check(int mid) {
        cnt.clear();
        inner.clear();
        p[0] = 1;
        for (int j = 0; j < paths.length; ++j) {
            int n = paths[j].length;
            for (int i = 1; i <= n; ++i) {
                p[i] = p[i - 1] * 133331;
                h[i] = h[i - 1] * 133331 + paths[j][i - 1];
            }
            for (int i = mid; i <= n; ++i) {
                long val = get(i - mid + 1, i);
                if (!inner.containsKey(val) || inner.get(val) != j) {
                    inner.put(val, j);
                    cnt.put(val, cnt.getOrDefault(val, 0) + 1);
                }
            }
        }
        int max = 0;
        for (int val : cnt.values()) {
            max = Math.max(max, val);
        }
        return max == paths.length;
    }

    private long get(int l, int r) {
        return h[r] - h[l - 1] * p[r - l + 1];
    }
}

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