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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2603\. Collect Coins in a Tree
Hard
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:
* Collect all the coins that are at a distance of at most `2` from the current vertex, or
* Move to any adjacent vertex in the tree.
Find _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.length`
* 1 <= n <= 3 * 104
* `0 <= coins[i] <= 1`
* `edges.length == n - 1`
* `edges[i].length == 2`
* 0 <= ai, bi < n
* ai != bi
* `edges` represents a valid tree.