Connect Sticks - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers sticks. 
Each element in the list represents a stick with two ends, and has two numbers between [1, 6] representing each end. 
You can connect two sticks together if any of their ends are equal. The resulting stick's ends become the leftover ends and its length is incremented.

Return the length of the longest stick possible.

Constraints

n ≤ 10 where n is the length of sticks.

Example 1

Input

sticks = [
    [1, 2],
    [1, 3],
    [2, 4],
    [6, 6]
]

Output

3

Explanation

We can connect [1, 2] and [1, 3] to get [2, 3], which we can connect it with [2, 4] to get [3, 4].


Solution :



title-img



                        Solution in C++ :

int ans = 0, n;
vector<vector<int>> adj;
vector<bool> vis;
void dfs(int u, int d) {
    vis[u % n] = true;
    vis[u % n + n] = true;
    ans = max(ans, d);
    for (int v : adj[u]) {
        if (vis[v]) continue;
        dfs(v, d + 1);
    }
    vis[u % n] = false;
    vis[u % n + n] = false;
}

int solve(vector<vector<int>>& sticks) {
    n = sticks.size();
    for (int i = 0; i < n; ++i) sticks.push_back({sticks[i][1], sticks[i][0]});
    int n2 = n << 1;
    adj = vector<vector<int>>(n2);
    for (int i = 0; i < n2; ++i)
        for (int j = 0; j < n2; ++j) {
            if (i == j || abs(i - j) == n) continue;
            if (sticks[i][1] == sticks[j][0]) adj[i].emplace_back(j);
        }
    for (int i = 0; i < n2; ++i) {
        vis.assign(n2, false);
        dfs(i, 1);
    }
    return ans;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] sticks) {
        return backtrack(sticks, null, new HashSet<>());
    }

    public int backtrack(int[][] sticks, int[] curr, Set<Integer> visited) {
        int ans = 0;

        for (int j = 0; j < sticks.length; j++) {
            if (visited.contains(j)) {
                continue;
            }

            visited.add(j);

            if (curr == null) {
                ans = Integer.max(
                    ans, 1 + backtrack(sticks, new int[] {sticks[j][0], sticks[j][1]}, visited));
            } else if (sticks[j][0] == curr[0]) {
                ans = Integer.max(
                    ans, 1 + backtrack(sticks, new int[] {sticks[j][1], curr[1]}, visited));
            } else if (sticks[j][1] == curr[0]) {
                ans = Integer.max(
                    ans, 1 + backtrack(sticks, new int[] {sticks[j][0], curr[1]}, visited));
            } else if (sticks[j][0] == curr[1]) {
                ans = Integer.max(
                    ans, 1 + backtrack(sticks, new int[] {sticks[j][1], curr[0]}, visited));
            } else if (sticks[j][1] == curr[1]) {
                ans = Integer.max(
                    ans, 1 + backtrack(sticks, new int[] {sticks[j][0], curr[0]}, visited));
            }

            visited.remove(j);
        }

        return ans;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, s):
        res = 0
        vis = set()
        # dfs = []
        def go(i, d, cur):
            nonlocal res, vis
            vis.add(i)
            # dfs.append(cur)
            # print(cur)
            for j in range(len(s)):
                if j in vis:
                    continue
                # chain em'
                if s[j][0] == cur[0]:
                    go(j, d + 1, [s[j][1], cur[1]])
                if s[j][1] == cur[0]:
                    go(j, d + 1, [s[j][0], cur[1]])
                if s[j][0] == cur[1]:
                    go(j, d + 1, [s[j][1], cur[0]])
                if s[j][1] == cur[1]:
                    go(j, d + 1, [s[j][0], cur[0]])
            vis.remove(i)
            res = max(res, d)
            # print(dfs, d)
            # dfs.pop()

        for i in range(len(s)):
            go(i, 1, s[i])

        return res
                    

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