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

Solution: Binary Search + Rolling Hash

• Time complexity: O(mnlogm)
• Space complexity: O(mn)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} paths
 * @return {number}
 */
const longestCommonSubpath = function (n, paths) {
  const kBase = 165131n;
  const kHash = 8417508174513n;
  let left = 0;
  let right = Math.min(...paths.map(path => path.length));

  const rollingHash = (path, mid) => {
    const m = path.length;
    const hashSet = new Set();
    let hash = 0n;
    let maxPower = 1n;

    for (let index = 0; index < mid; index++) {
      const country = BigInt(path[index]);

      hash = (hash * kBase + country) % kHash;

      if (index < mid - 1) {
        maxPower = (maxPower * kBase) % kHash;
      }
    }

    hashSet.add(hash);

    for (let index = mid; index < m; index++) {
      const country = BigInt(path[index]);
      const leftCountry = BigInt(path[index - mid]);

      hash = (hash - ((leftCountry * maxPower) % kHash) + kHash) % kHash;
      hash = (hash * kBase + country) % kHash;
      hashSet.add(hash);
    }

    return hashSet;
  };

  const commonSubpath = mid => {
    const hashSets = paths.map(path => rollingHash(path, mid));

    for (const hash of hashSets[0]) {
      if (hashSets.every(set => set.has(hash))) {
        return true;
      }
    }

    return false;
  };

  while (left <= right) {
    const mid = Math.floor((left + right) / 2);

    commonSubpath(mid) ? (left = mid + 1) : (right = mid - 1);
  }

  return right;
};