There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given an integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given an array coins of size n where coins[i] can be either 0 or 1, where 1 indicates the presence of a coin in the vertex i.
Initially, you choose to start at any vertex in the tree. Then, you can perform the following operations any number of times:
2 from the current vertex, orFind the minimum number of edges you need to go through to collect all the coins and go back to the initial vertex.
Note that if you pass an edge several times, you need to count it into the answer several times.
Example 1:
Input: coins = [1,0,0,0,0,1], edges = [[0,1],[1,2],[2,3],[3,4],[4,5]] Output: 2 Explanation: Start at vertex 2, collect the coin at vertex 0, move to vertex 3, collect the coin at vertex 5 then move back to vertex 2.
Example 2:
Input: coins = [0,0,0,1,1,0,0,1], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[5,6],[5,7]] Output: 2 Explanation: Start at vertex 0, collect the coins at vertices 4 and 3, move to vertex 2, collect the coin at vertex 7, then move back to vertex 0.
Constraints:
n == coins.length1 <= n <= 3 * 1040 <= coins[i] <= 1edges.length == n - 1edges[i].length == 20 <= ai, bi < nai != biedges represents a valid tree.This approach uses a DFS traversal to determine the minimal path needed to return to the starting node after collecting all coins. It calculates the number of edges that need to be traversed with consideration of reachable coins at each step. The DFS explores each node and checks its adjacent nodes for coin presence alongside setting an optimal route back.
The C solution implements a DFS function to traverse the graph. The configuration uses adjacency lists to represent the edges of the tree. By invoking dfs(), it calculates the total number of unique paths required to collect all coins and return to the starting node. The function collectCoins() sums two times the DFS result minus one, owing to counting each edge twice on the return trip.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n) due to the DFS traversal visiting all nodes once.
Space Complexity: O(n) for the adjacency list representation and visited array.
This approach involves using the Breadth-First Search (BFS) method with a queue to dynamically explore the tree in slices of depth-limited exploration. It records each potential realization of coin acquisition using a range-bound periphery, leveraging spatial layers to evaluate all reachable nodes distinctly within optimal edge traversal focus.
The BFS solution in C utilizes an adjacency matrix to represent connections between nodes after initializing levels and visiting indices. The breadth-first circuit run systematically identifies reachability and collates edge quantity necessary for traversing given coin presence.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n^2) due to the adjacency matrix and bilinear node checks.
Space Complexity: O(n^2) in maintaining matrix overhead structure.
| Approach | Complexity |
|---|---|
| Depth-First Search (DFS) Approach | Time Complexity: O(n) due to the DFS traversal visiting all nodes once. |
| BFS and Queue-Based Approach | Time Complexity: O(n^2) due to the adjacency matrix and bilinear node checks. |
Coin Change - Dynamic Programming Bottom Up - Leetcode 322 • NeetCode • 574,201 views views
Watch 9 more video solutions →Practice Collect Coins in a Tree with our built-in code editor and test cases.
Practice on FleetCode