Collect Maximum Points from all Nodes

Given a tree of N nodes and N-1 edges rooted at Node 1, an array points[] of size N, where points[i] denote the number of points at the vertex i and an integer k. A Node can be visited only if its ancestors have already been visited. When any Node i is visited 2 types of operations can be applied:

• Type 1: Add (points[i] - k) points to total points.
• Type 2: Add floor(points[i] / 2) points to total points and all the nodes j in the subtree of node i becomes floor(coins[j] / 2).

The task is to determine the maximum points that can be collected from all nodes.

Constraints:

• 2 <= N <= 10^5
• 0 <= points [i] <= 10^9
• 0 <= x <= 10^9

Examples:

Input: N=5, K=4

Collect Maximum Points from all Nodes

Output: 10
Explanation:

• Type 1 operation is performed on Node 1, so points obtained = (7 – 4) = 3
• Type 1 operation is performed on Node 3, so points obtained = (9 – 4) = 5
• Type 2 operation is performed on Node 4, so points obtained = floor(5/2) = 2 and since Node 4 has no nodes in the subtree, therefore no other node’s value gets reduced.
• Type 2 operation is performed on Node 2, so points obtained = floor(0/2) = 0 and since Node 5 lies in the subtree of node 2, value of node 5 gets reduced to floor(3/2) = 1.
• Type 2 operation is performed on Node 5, so points obtained = floor(1/2) = 0 and since Node 5 has no nodes in the subtree, therefore no other node’s value gets reduced.

Total points obtained = 3 + 5 + 2 + 0 + 0 = 10

Input: N=3, K=0

Collect Maximum Points from all Nodes

Output: 7
Explanation:

• Type 1 operation is performed in Node 1, so points obtained = (2 – 0) = 2
• Type 1 operation is performed in Node 2, so points obtained = (2 – 0) = 2
• Type 1 operation is performed in Node 3, so points obtained = (3 – 0) = 3

Total points obtained = 2 + 2 + 3 = 7

Collect Maximum Points from all Nodes using Dynamic Programming:

Approach: The problem can be solved using the following approach:

It can be observed that operation of type 2 can be applied at most 30 times because each type 2 operation reduces the value points at a node i to points[i]/2 and if 30 operations of type 2 are applied then node with any value up to 10^9 will get reduced to 0 as 2³⁰>10⁹ so type 2 operation will be applied at most 30 times. Now we can apply Dynamic Programming to solve this problem. The state will be: dp[curr][i] = maximum points that can be collected at the subtree rooted at node curr such that type 2 operation are already applied i times on the ancestors till now. The answer will be max(dp[1][0], dp[1][1], dp[1][2], …… , dp[1][30]).

Follow the steps to solve the problem:

• Create a 2D dp[][] array of size N X 30 , where dp[curr][i] represents maximum points that can be collected at the subtree rooted at node curr with type 2 operation already applied i times on the ancestors till now.
• Iterate through the nodes using DFS. At any node curr, dp[curr][i] will be calculated by:
  • Applying Type 1 Operation: Let sum1 denote the maximum points that can be collected from the subtree rooted at node curr if type 1 operation is performed on node curr. So sum1 will be:
    , where x is child of node curr and pt is the points at node curr when operation 2 have applied i times before visiting node curr.
  • Applying Type 2 Operation: Let sum2 denote the maximum points that can be collected from the subtree rooted at node curr if type 2 operation is performed on node curr. So sum2 will be:
    , where x is child of node curr and pt is the points at node curr when operation 2 have applied i times before visiting node curr.
  • dp[curr][i] will be max(sum1, sum2).
• The final answer will be max(dp[1][0], dp[1][1], ……, dp[1][29], dp[1][30]).

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Depth-First Search (DFS) function to process the tree
void dfs(vector<vector<int> >& adj, vector<int>& points,
         vector<vector<int> >& dp, int k, int curr, int par)
{
    // Traverse the children nodes of the current node
    for (auto x : adj[curr]) {
        if (par != x) {
            dfs(adj, points, dp, k, x, curr);
        }
    }

    int pt = points[curr - 1];

    // Loop for 30 iterations, corresponding to applying
    // type 2 operation at most 30 times
    for (int i = 0; i < 30; i++) {
        int sum2 = pt / 2, sum1 = pt - k;

        // Iterate through the child nodes
        for (auto x : adj[curr]) {
            if (x != par) {
                sum2 = sum2 + dp[x][i + 1];
                sum1 = sum1 + dp[x][i];
            }
        }

        pt = pt / 2;

        // Update the DP values for the current node and
        // iteration
        dp[curr][i] = max(sum1, sum2);
    }
}

// Function to calculate the maximum points
int maximumPoints(vector<vector<int> >& edges,
                  vector<int>& points, int k)
{
    int n = points.size();
    vector<vector<int> > adj(n + 1);

    // Create an adjacency list based on the input edges
    for (int i = 0; i < edges.size(); i++) {
        adj[edges[i][0]].push_back(edges[i][1]);
        adj[edges[i][1]].push_back(edges[i][0]);
    }

    // Initialize a 2D DP array with zeros
    vector<vector<int> > dp(n + 1, vector<int>(31, 0));

    // Perform DFS to calculate DP values
    dfs(adj, points, dp, k, 1, -1);

    int ans = 0;

    // Find the maximum points for all iterations
    for (int i = 0; i < 31; i++) {
        ans = max(ans, dp[1][i]);
    }

    return ans;
}

// Driver Code
int main()
{
    int N = 5;
    int k = 4;
    vector<vector<int> > edges
        = { { 1, 2 }, { 1, 3 }, { 1, 4 }, { 2, 5 } };
    vector<int> points = { 7, 0, 9, 5, 3 };
    cout << maximumPoints(edges, points, k);
}

Java

import java.util.*;
// Nikunj Sonigara
class Main {

    // Depth-First Search (DFS) function to process the tree
    static void dfs(List<List<Integer>> adj, List<Integer> points,
                    List<List<Integer>> dp, int k, int curr, int par) {
        // Traverse the children nodes of the current node
        for (int x : adj.get(curr)) {
            if (par != x) {
                dfs(adj, points, dp, k, x, curr);
            }
        }

        int pt = points.get(curr - 1);

        // Loop for 30 iterations, corresponding to applying
        // type 2 operation at most 30 times
        for (int i = 0; i < 30; i++) {
            int sum2 = pt / 2, sum1 = pt - k;

            // Iterate through the child nodes
            for (int x : adj.get(curr)) {
                if (x != par) {
                    sum2 = sum2 + dp.get(x).get(i + 1);
                    sum1 = sum1 + dp.get(x).get(i);
                }
            }

            pt = pt / 2;

            // Update the DP values for the current node and
            // iteration
            dp.get(curr).set(i, Math.max(sum1, sum2));
        }
    }

    // Function to calculate the maximum points
    static int maximumPoints(List<List<Integer>> edges,
                             List<Integer> points, int k) {
        int n = points.size();
        List<List<Integer>> adj = new ArrayList<>(n + 1);

        // Initialize adjacency list
        for (int i = 0; i <= n; i++) {
            adj.add(new ArrayList<>());
        }

        // Create an adjacency list based on the input edges
        for (List<Integer> edge : edges) {
            adj.get(edge.get(0)).add(edge.get(1));
            adj.get(edge.get(1)).add(edge.get(0));
        }

        // Initialize a 2D DP array with zeros
        List<List<Integer>> dp = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            dp.add(new ArrayList<>(31));
            for (int j = 0; j < 31; j++) {
                dp.get(i).add(0);
            }
        }

        // Perform DFS to calculate DP values
        dfs(adj, points, dp, k, 1, -1);

        int ans = 0;

        // Find the maximum points for all iterations
        for (int i = 0; i < 31; i++) {
            ans = Math.max(ans, dp.get(1).get(i));
        }

        return ans;
    }

    // Driver Code
    public static void main(String[] args) {
        int N = 5;
        int k = 4;
        List<List<Integer>> edges = List.of(
                List.of(1, 2),
                List.of(1, 3),
                List.of(1, 4),
                List.of(2, 5)
        );
        List<Integer> points = List.of(7, 0, 9, 5, 3);
        System.out.println(maximumPoints(edges, points, k));
    }
}

Python3

# Depth-First Search (DFS) function to process the tree
def dfs(adj, points, dp, k, curr, par):
    # Traverse the children nodes of the current node
    for x in adj[curr]:
        if par != x:
            dfs(adj, points, dp, k, x, curr)

    pt = points[curr - 1]

    # Loop for 30 iterations, corresponding to applying
    # type 2 operation at most 30 times
    for i in range(30):
        sum2, sum1 = pt // 2, pt - k

        # Iterate through the child nodes
        for x in adj[curr]:
            if x != par:
                sum2 += dp[x][i + 1]
                sum1 += dp[x][i]

        pt //= 2

        # Update the DP values for the current node and iteration
        dp[curr][i] = max(sum1, sum2)

# Function to calculate the maximum points
def maximumPoints(edges, points, k):
    n = len(points)
    adj = [[] for _ in range(n + 1)]

    # Create an adjacency list based on the input edges
    for edge in edges:
        adj[edge[0]].append(edge[1])
        adj[edge[1]].append(edge[0])

    # Initialize a 2D DP array with zeros
    dp = [[0] * 31 for _ in range(n + 1)]

    # Perform DFS to calculate DP values
    dfs(adj, points, dp, k, 1, -1)

    ans = 0

    # Find the maximum points for all iterations
    for i in range(31):
        ans = max(ans, dp[1][i])

    return ans

# Driver Code
def main():
    N = 5
    k = 4
    edges = [[1, 2], [1, 3], [1, 4], [2, 5]]
    points = [7, 0, 9, 5, 3]
    print(maximumPoints(edges, points, k))

if __name__ == "__main__":
    main()

C#

using System;
using System.Collections.Generic;

class MaximumPointsCalculator
{
    // Depth-First Search (DFS) function to process the tree
    static void Dfs(List<List<int>> adj, List<int> points,
                    List<List<int>> dp, int k, int curr, int par)
    {
        // Traverse the children nodes of the current node
        foreach (int x in adj[curr])
        {
            if (par != x)
            {
                Dfs(adj, points, dp, k, x, curr);
            }
        }

        int pt = points[curr - 1];

        // Loop for 30 iterations, corresponding to applying
        // type 2 operation at most 30 times
        for (int i = 0; i < 30; i++)
        {
            int sum2 = pt / 2, sum1 = pt - k;

            // Iterate through the child nodes
            foreach (int x in adj[curr])
            {
                if (x != par)
                {
                    sum2 = sum2 + dp[x][i + 1];
                    sum1 = sum1 + dp[x][i];
                }
            }

            pt = pt / 2;

            // Update the DP values for the current node and
            // iteration
            dp[curr][i] = Math.Max(sum1, sum2);
        }
    }

    // Function to calculate the maximum points
    static int MaximumPoints(List<List<int>> edges,
                             List<int> points, int k)
    {
        int n = points.Count;
        List<List<int>> adj = new List<List<int>>();

        // Initialize adjacency list
        for (int i = 0; i <= n; i++)
        {
            adj.Add(new List<int>());
        }

        // Create an adjacency list based on the input edges
        foreach (List<int> edge in edges)
        {
            adj[edge[0]].Add(edge[1]);
            adj[edge[1]].Add(edge[0]);
        }

        // Initialize a 2D DP array with zeros
        List<List<int>> dp = new List<List<int>>();
        for (int i = 0; i <= n; i++)
        {
            dp.Add(new List<int>(new int[31]));
        }

        // Perform DFS to calculate DP values
        Dfs(adj, points, dp, k, 1, -1);

        int ans = 0;

        // Find the maximum points for all iterations
        for (int i = 0; i < 31; i++)
        {
            ans = Math.Max(ans, dp[1][i]);
        }

        return ans;
    }

    // Driver Code
    static void Main(string[] args)
    {
        int k = 4;
        List<List<int>> edges = new List<List<int>>
        {
            new List<int> { 1, 2 },
            new List<int> { 1, 3 },
            new List<int> { 1, 4 },
            new List<int> { 2, 5 }
        };
        List<int> points = new List<int> { 7, 0, 9, 5, 3 };
        Console.WriteLine(MaximumPoints(edges, points, k));
    }
}

Javascript

function dfs(adj, points, dp, k, curr, par) {
    // Traverse the children nodes of current node
    for (let x of adj[curr]) {
        if (par !== x) {
            dfs(adj, points, dp, k, x, curr);
        }
    }
    let pt = points[curr - 1];
    // Loop for 30 iterations, corresponding to applying
    for (let i = 0; i < 30; i++) {
        let sum2 = Math.floor(pt / 2);
        let sum1 = pt - k;
        // Iterate through the child nodes
        for (let x of adj[curr]) {
            if (x !== par) {
                sum2 += dp[x][i + 1];
                sum1 += dp[x][i];
            }
        }
        pt = Math.floor(pt / 2);
        // Update the DP values for the current node and iteration
        dp[curr][i] = Math.max(sum1, sum2);
    }
}
// Function to calculate the maximum points
function GFG(edges, points, k) {
    const n = points.length;
    const adj = Array.from({ length: n + 1 }, () => []);
    // Create an adjacency list based on input edges
    for (let i = 0; i < edges.length; i++) {
        adj[edges[i][0]].push(edges[i][1]);
        adj[edges[i][1]].push(edges[i][0]);
    }
    // Initialize a 2D DP array with the zeros
    const dp = Array.from({ length: n + 1 }, () => Array(31).fill(0));
    dfs(adj, points, dp, k, 1, -1);
    let ans = 0;
    // Find the maximum points for the all iterations
    for (let i = 0; i < 31; i++) {
        ans = Math.max(ans, dp[1][i]);
    }
    return ans;
}
// Driver Code
const N = 5;
const k = 4;
const edges = [[1, 2], [1, 3], [1, 4], [2, 5]];
const points = [7, 0, 9, 5, 3];
console.log(GFG(edges, points, k));

Output

10

Time Complexity: O(N), where N is number of nodes in the tree
Auxiliary Space: O(N)