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]
.
class Solution:
def longestCommonSubpath(self, n: int, paths: List[List[int]]) -> int:
def get(l, r, h):
return (h[r] - h[l - 1] * p[r - l + 1]) % mod
def check(l):
cnt = Counter()
for k, path in enumerate(paths):
vis = set()
for i in range(len(path) - l + 1):
j = i + l - 1
x = get(i + 1, j + 1, hh[k])
if x not in vis:
vis.add(x)
cnt[x] += 1
return max(cnt.values()) == len(paths)
base = 133331
mod = 2**64 + 1
p = [0] * 100010
p[0] = 1
for i in range(1, len(p)):
p[i] = (p[i - 1] * base) % mod
hh = []
for path in paths:
h = [0] * (len(path) + 10)
for j, c in enumerate(path):
h[j + 1] = (h[j] * base) % mod + c
hh.append(h)
left, right = 0, min(len(path) for path in paths)
while left < right:
mid = (left + right + 1) >> 1
if check(mid):
left = mid
else:
right = mid - 1
return left
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];
}
}