Longest Common Subpath Leetcode Solution

Longest Common Subpath

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.

See Also: Weekly Contest 248

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.

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].

Solution

Program C++

Credit: HERE

Please note that we have to choose base wisely i.e. base >= 100007. Otherwise it’s just a simple rolling hash question. Also, the constraints of the problem were misleading during the contest.

#define ll long long

class Solution {
public:
    
    ll base, MOD;
    vector<ll> p;
    
    bool works(ll len, vector<vector<int>>& paths, ll m){
        unordered_set<ll> s;
        for(ll i=0; i<m; i++){
            if(paths[i].size()<len){
                return false;
            }
            unordered_set<ll> ss;
            ll hash=0;
            for(ll j=0; j<len; j++){
                hash=(hash+(((ll)(paths[i][j]+1)*p[len-1-j])%MOD))%MOD;
            }
            if(i==0){
                s.insert(hash);
            }
            else{
                if(s.find(hash)!=s.end()){
                    ss.insert(hash);
                }
            }
            for(ll j=len; j<paths[i].size(); j++){
                hash=(hash+MOD-(((ll)(paths[i][j-len]+1)*p[len-1])%MOD))%MOD;
                hash=(ll)hash*base%MOD;
                hash=(hash+paths[i][j]+1)%MOD;
                if(i==0){
                    s.insert(hash);
                }
                else{
                    if(s.find(hash)!=s.end()){
                        ss.insert(hash);
                    }
                }
            }
            if(i>0){
                s=ss;
            }
            if(s.size()==0){
                return false;
            }
        }
        if(s.size()>0){
            return true;
        }
        else{
            return false;
        }
    }
    
    int longestCommonSubpath(int n, vector<vector<int>>& paths) {
        ll l=0, r=1e9L, m=paths.size();
        base=100007, MOD=1e11L+7;
        for(ll i=0; i<m; i++){
            r=min(r, (ll)paths[i].size());
        }
        p=vector<ll>(r);
        p[0]=1;
        for(ll i=1; i<r; i++){
            p[i]=(ll)base*p[i-1]%MOD;
        }
        while(l<r){
            ll mid=(l+r+1)/2;
            if(works(mid, paths, m)){
                l=mid;
            }
            else{
                r=mid-1;
            }
        }
        return r;
    }
};

Program Python

class Solution:
    def longestCommonSubpath(self, n: int, paths: List[List[int]]) -> int:
        min_path = math.inf
        for p in paths:
            min_path = min(len(p), min_path)
            
        def possible(guess):
            index = Counter()
            for p in paths:
                seen = set()
                for i in range(len(p)-guess+1):
                    key = tuple(p[i:i+guess])
                    if key not in seen:
                        index[key] += 1
                        seen.add(key)
                        
                        if index[key] == len(paths):
                            return True
            return False

        lo, hi = 0, min_path
        while lo < hi:
            guess = (lo + hi + 1) // 2
            if possible(guess):
                lo = guess
            else:
                hi = guess - 1
        return lo

Program Java

Credit: HERE

Build a trie in n^2 time using the smallest path. Then, test against the trie with the other paths in n^2 time each. Probably wont get below the time limit with this approach, but it is a different way to approach other than hashing.

class Solution {
    public int longestCommonSubpath(int n, int[][] paths) {
        TrieNode trie = new TrieNode();
        int longestFull = 0;
        int shortestPathNum = 0;
        int shortestPathLength = paths[0].length;
        for (int i = 0; i < paths.length; i++) {
            if (paths[i].length < shortestPathLength) {
                shortestPathLength = paths[i].length;
                shortestPathNum = i;
            }
        }
        
        // build Trie
        for (int i = 0; i < paths[shortestPathNum].length; i++) {
            TrieNode tempTrieNode = trie;
            for (int j = i; j < paths[shortestPathNum].length; j++) {
                int city = paths[shortestPathNum][j];
                if (!tempTrieNode.next.containsKey(city)) tempTrieNode.next.put(city, new TrieNode());
                TrieNode nextNode = tempTrieNode.next.get(city);
                nextNode.pathsSeen.add(shortestPathNum);
                tempTrieNode = nextNode;
            }
        }
        
        // test Trie
        for (int pathNum = 0; pathNum < paths.length; pathNum++) {
            if (pathNum == shortestPathNum) continue; // no need to test the builder
            int[] path = paths[pathNum];
            for (int i = 0; i < path.length; i++) {
                TrieNode tempTrieNode = trie;
                for (int j = i; j < path.length; j++) {
                    int city = path[j];
                    if (!tempTrieNode.next.containsKey(city)) break;
                    TrieNode nextNode = tempTrieNode.next.get(city);
                    if (nextNode.pathsSeen.size() < pathNum) break;
                    nextNode.pathsSeen.add(pathNum);
                    if (nextNode.pathsSeen.size() == paths.length) {
                        longestFull = Math.max(longestFull, j - i + 1);
                    }
                    tempTrieNode = nextNode;
                }
            }
        }
        return longestFull;
    }
    
    class TrieNode {
        public Map<Integer, TrieNode> next = new HashMap<>();
        public Set<Integer> pathsSeen = new HashSet<>();
    }
}

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