Graph - Medium Level - Question 1

Graph - Medium Level - Question 1

Leetcode 2049. Count Nodes With the Highest Score

There is a binary tree rooted at 0 consisting of n nodes. The nodes are labeled from 0 to n - 1. You are given a 0-indexed integer array parents representing the tree, where parents[i] is the parent of node i. Since node 0 is the root, parents[0] == -1.

Each node has a score. To find the score of a node, consider if the node and the edges connected to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of the node is the product of the sizes of all those subtrees.

Return the number of nodes that have the highest score.

Constraints:

n == parents.length

2 <= n <= 10^5

parents[0] == -1

0 <= parents[i] <= n - 1 for i != 0

parents represents a valid binary tree.

Analysis:

If we have had the binary tree, then we just can do a top-down count, to count the number of nodes for the sub-tree with the root as the current nodes. But we do not have the tree, instead, we only know the parents of each nodes.

So we need to construct the binary tree by ourselves. This process can be generalized to build a graph.

Since we need to know the number of nodes of the subtree with the root as the current node, so we need to know the children nodes of each nodes. And also we need to know the root.

For this question, we can use a two dimensional array to simply the graph notation. The array index would be the node index, and the array contents are for the children nodes.

After having the graph, we can do a top-down traversal to count the number of nodes "below" the node, starting with the root node. A dynamical programming trick is needed here as well, to record the numbers.

After having all the information we need, we can check each node, to see which one give the highest score, and count its times. If the node is the root, we just need to check its children nodes.

See the code below:

class Solution {
public:
    int countHighestScoreNodes(vector<int>& parents) {
        int n = parents.size(), root = 0;
        long res = -1, num = 0;
        vector<long> dp(n, 0);
        vector<vector<long>> g(n);
        for(int i=0; i<n; ++i) {
            if(parents[i] == -1) root = i;
            else g[parents[i]].push_back(i);
        }
        ct(root, g, dp);
        for(int i=0; i<n; ++i) {
            // cout<<i<<" "<<dp[i]<<endl;
        }
        for(int i=0; i<n; ++i) {
            long one = 0;
            if(i == root) {
                if(g[i].size() == 2) one = dp[g[i][0]]*dp[g[i][1]];
                else if(g[i].size() == 1) one = dp[g[i][0]];
            }
            else {
                if(g[i].size() == 0) one = dp[root] - 1;
                else if(g[i].size() == 1) one = dp[g[i][0]] *(dp[root] - dp[i]);
                else if(g[i].size() == 2) one = dp[g[i][0]] * dp[g[i][1]] * (dp[root] - dp[i]);
            }
            if(res < one) {
                res = one;
                num = 1;
            }
            else if(res == one) ++num;
        }
        return num;
    }
private:
    long ct(int root, vector<vector<long>>& g, vector<long>& dp) {
        for(auto &a : g[root]) {
            dp[root] += ct(a, g, dp);
        }
        return dp[root] += 1;
    }
};

Made some change to let the code can be used for a n-ary tree:

class Solution {
public:
    int countHighestScoreNodes(vector<int>& parents) {
        int n = parents.size(), root = 0;
        long res = -1, num = 0;
        vector<long> dp(n, 0);
        vector<vector<long>> g(n);
        for(int i=0; i<n; ++i) {
            if(parents[i] == -1) root = i;
            else g[parents[i]].push_back(i);
        }
        ct(root, g, dp);
        for(int i=0; i<n; ++i) {
            long one = i == root ? 1 : dp[root] - dp[i];
            for(auto &a : g[i]) {
                one *= dp[a];
            }
            if(res < one) {
                res = one;
                num = 1;
            }
            else if(res == one) ++num;
        }
        return num;
    }
private:
    long ct(int root, vector<vector<long>>& g, vector<long>& dp) {
        for(auto &a : g[root]) {
            dp[root] += ct(a, g, dp);
        }
        return dp[root] += 1;
    }
};

Upper Layer