# 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 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

• 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:
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