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1923. Longest Common Subpath
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 i^{th}
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 <= 10^{5}
m == paths.length
2 <= m <= 10^{5}
sum(paths[i].length) <= 10^{5}
0 <= paths[i][j] < n
 The same city is not listed multiple times consecutively in
paths[i]
.
Solutions

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]; } }

class Solution: def longestCommonSubpath(self, n: int, paths: List[List[int]]) > int: def check(k: int) > bool: cnt = Counter() for h in hh: vis = set() for i in range(1, len(h)  k + 1): j = i + k  1 x = (h[j]  h[i  1] * p[j  i + 1]) % mod if x not in vis: vis.add(x) cnt[x] += 1 return max(cnt.values()) == m m = len(paths) mx = max(len(path) for path in paths) base = 133331 mod = 2**64 + 1 p = [0] * (mx + 1) p[0] = 1 for i in range(1, len(p)): p[i] = p[i  1] * base % mod hh = [] for path in paths: k = len(path) h = [0] * (k + 1) for i, x in enumerate(path, 1): h[i] = h[i  1] * base % mod + x hh.append(h) l, r = 0, min(len(path) for path in paths) while l < r: mid = (l + r + 1) >> 1 if check(mid): l = mid else: r = mid  1 return l