LeetCode 2973: Find Number of Coins to Place in Tree Nodes Solution

Master LeetCode problem 2973 (Find Number of Coins to Place in Tree Nodes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2973. Find Number of Coins to Place in Tree Nodes

Hard

Dynamic Programming Tree Depth-First Search Sorting Heap (Priority Queue)

Problem Explanation

Explanation:

To solve this problem, we can use a depth-first search (DFS) approach to recursively calculate the number of coins to be placed at each node. We will keep track of the subtree size and the maximum product of cost values for 3 distinct nodes in each subtree.

Define a recursive function dfs that takes the current node u and its parent parent as parameters.
Initialize the subtree size subtreeSize[u] to 1.
For each child v of node u, recursively call dfs(v, u) and update subtreeSize[u] by adding subtreeSize[v].
If subtreeSize[u] < 3, place 1 coin at node u.
Otherwise, calculate the maximum product of cost values for 3 distinct nodes in the subtree of node u.
Return the maximum product of cost values for node u.
The result will be the number of coins placed at each node. :

Solution Code

class Solution {
    int[] subtreeSize;
    int[] coins;
    List<Integer>[] tree;
    int[] cost;
    
    public int[] distributeCoins(int[][] edges, int[] cost) {
        int n = cost.length;
        this.cost = cost;
        subtreeSize = new int[n];
        coins = new int[n];
        tree = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            tree[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            tree[edge[0]].add(edge[1]);
            tree[edge[1]].add(edge[0]);
        }
        dfs(0, -1);
        return coins;
    }
    
    private int dfs(int u, int parent) {
        subtreeSize[u] = 1;
        for (int v : tree[u]) {
            if (v == parent) continue;
            subtreeSize[u] += dfs(v, u);
        }
        if (subtreeSize[u] < 3) {
            coins[u] = 1;
            return subtreeSize[u];
        }
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Comparator.reverseOrder());
        for (int v : tree[u]) {
            if (v == parent) continue;
            maxHeap.offer(dfs(v, u));
        }
        int product = 1;
        for (int i = 0; i < 3; i++) {
            product *= maxHeap.poll();
        }
        coins[u] = Math.max(0, product * cost[u]);
        return subtreeSize[u];
    }
}

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