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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 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

• Java
• Python

• 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
  
  

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