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Traversal of tree with k jumps allowed between nodes of same height

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  • Difficulty Level : Expert
  • Last Updated : 31 Oct, 2022
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There is a tree with N nodes and node 1 is the root node. Each node of the tree can contain fruit or not. Initially, you are at the root node and start climbing on the tree. You can jump from a node to any node at the same level(i.e. the height of nodes from the root are same), During climbing from root node you can only make maximum K jumps. (K < 20) Now you have to climb on the tree (from root node-> any leaf node) in such a way so that you can collect maximum no of fruits.

Example :

Input Tree
Number of Nodes N = 12
Number of jumps allowed : 2
Edges:
1 2
1 3
2 4
2 5
5 9
9 10
9 11
11 12
3 7
7 6
7 8
no of node having fruit(nf) : 8
Nodes Containing Fruits(lvn) : 2 4 5 7 8 9 11 12
Output: 7

Tree for above testcase : 

 Explanation: 

Approach: The idea is to use DFS to create a Height Adjacency List of the Nodes and to store the parents. Then use another dfs to compute the maximum no of special nodes that can be reached using the following dp state:

dp[current_node][j] = max( max{ dp[child_i][j], for all children of current_node },
                         max{ dp[node_at_same_height_i][j - 1],
                         for all nodes at same height as current_node} )

Thus, dp[Root_Node][Total_no_of_Jumps] gives the answer to the problem. 

Below is the implementation of the above approach : 

CPP




// Program to demonstrate tree traversal with
// ability to jump between nodes of same height
#include <bits/stdc++.h>
using namespace std;
 
#define N 1000
 
vector<int> H[N];
 
// Arrays declaration
int Fruit[N];
int Parent[N];
int dp[N][20];
 
// Function for DFS
void dfs1(vector<int> tree[], int s, int p, int h)
{
    Parent[s] = p;
    int i;
    H[h].push_back(s);
    for (i = 0; i < tree[s].size(); i++) {
        int v = tree[s][i];
        if (v != p)
            dfs1(tree, v, s, h + 1);
    }
}
 
// Function for DFS
int dfs2(vector<int> tree[], int s, int p, int h, int j)
{
    int i;
    int ans = 0;
    if (dp[s][j] != -1)
        return dp[s][j];
 
    // jump
    if (j > 0) {
        for (i = 0; i < H[h].size(); i++) {
            int v = H[h][i];
            if (v != s)
                ans = max(ans, dfs2(tree, v, Parent[v], h,
                                    j - 1));
        }
    }
 
    // climb
    for (i = 0; i < tree[s].size(); i++) {
        int v = tree[s][i];
        if (v != p)
            ans = max(ans, dfs2(tree, v, s, h + 1, j));
    }
 
    if (Fruit[s] == 1)
        ans++;
    dp[s][j] = ans;
 
    return ans;
}
 
// Function to calculate and
// return maximum number of fruits
int maxFruit(vector<int> tree[], int NodesWithFruits[],
             int n, int m, int k)
{
    // reseting dp table and Fruit array
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < 20; j++)
            dp[i][j] = -1;
        Fruit[i] = 0;
    }
 
    // This array is used to mark
    // which nodes contain Fruits
    for (int i = 0; i < m; i++)
        Fruit[NodesWithFruits[i]] = 1;
 
    dfs1(tree, 1, 0, 0);
    int ans = dfs2(tree, 1, 0, 0, k);
 
    return ans;
}
 
// Function to add Edge
void addEdge(vector<int> tree[], int u, int v)
{
    tree[u].push_back(v);
    tree[v].push_back(u);
}
 
// Driver Code
int main()
{
    int n = 12; // Number of nodes
    int k = 2; // Number of allowed jumps
 
    vector<int> tree[N];
 
    // Edges
    addEdge(tree, 1, 2);
    addEdge(tree, 1, 3);
    addEdge(tree, 2, 4);
    addEdge(tree, 2, 5);
    addEdge(tree, 5, 9);
    addEdge(tree, 9, 10);
    addEdge(tree, 9, 11);
    addEdge(tree, 11, 12);
    addEdge(tree, 3, 7);
    addEdge(tree, 7, 6);
    addEdge(tree, 7, 8);
 
    int NodesWithFruits[] = { 2, 4, 5, 7, 8, 9, 11, 12 };
 
    // Number of nodes with fruits
    int m = sizeof(NodesWithFruits)
            / sizeof(NodesWithFruits[0]);
 
    int ans = maxFruit(tree, NodesWithFruits, n, m, k);
 
    cout << ans << endl;
 
    return 0;
}


Java




// Program to demonstrate tree traversal with
// ability to jump between nodes of same height
import java.util.*;
 
public class GFG {
    static int N;
 
    static ArrayList<ArrayList<Integer> > H;
 
    // Arrays declaration
    static int[] Fruit;
    static int[] Parent;
    static int[][] dp; //[N][20]
 
    // Function for DFS
    static void dfs1(ArrayList<ArrayList<Integer> > tree,
                     int s, int p, int h)
    {
        Parent[s] = p;
        int i;
        H.get(h).add(s);
        for (i = 0; i < tree.get(s).size(); i++) {
            int v = tree.get(s).get(i);
            if (v != p)
                dfs1(tree, v, s, h + 1);
        }
    }
 
    // Function for DFS
    static int dfs2(ArrayList<ArrayList<Integer> > tree,
                    int s, int p, int h, int j)
    {
        int i;
        int ans = 0;
        if (dp[s][j] != -1)
            return dp[s][j];
 
        // jump
        if (j > 0) {
            for (i = 0; i < H.get(h).size(); i++) {
                int v = H.get(h).get(i);
                if (v != s)
                    ans = Math.max(
                        ans,
                        dfs2(tree, v, Parent[v], h, j - 1));
            }
        }
 
        // climb
        for (i = 0; i < tree.get(s).size(); i++) {
            int v = tree.get(s).get(i);
            if (v != p)
                ans = Math.max(ans,
                               dfs2(tree, v, s, h + 1, j));
        }
 
        if (Fruit[s] == 1)
            ans++;
        dp[s][j] = ans;
 
        return ans;
    }
 
    // Function to calculate and
    // return maximum number of fruits
    static int maxFruit(ArrayList<ArrayList<Integer> > tree,
                        int NodesWithFruits[], int n, int m,
                        int k)
    {
        // reseting dp table and Fruit array
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < 20; j++)
                dp[i][j] = -1;
            Fruit[i] = 0;
        }
 
        // This array is used to mark
        // which nodes contain Fruits
        for (int i = 0; i < m; i++)
            Fruit[NodesWithFruits[i]] = 1;
 
        dfs1(tree, 1, 0, 0);
        int ans = dfs2(tree, 1, 0, 0, k);
 
        return ans;
    }
 
    // Function to add Edge
    static void addEdge(ArrayList<ArrayList<Integer> > tree,
                        int u, int v)
    {
        tree.get(u).add(v);
        tree.get(v).add(u);
    }
 
    public static void main(String[] args)
    {
        N = 1000;
        H = new ArrayList<>();
        Fruit = new int[N];
        Parent = new int[N];
        dp = new int[N][20];
 
        int n = 12; // Number of nodes
        int k = 2; // Number of allowed jumps
 
        ArrayList<ArrayList<Integer> > tree
            = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            tree.add(new ArrayList<>());
            H.add(new ArrayList<>());
        }
        // Edges
        addEdge(tree, 1, 2);
        addEdge(tree, 1, 3);
        addEdge(tree, 2, 4);
        addEdge(tree, 2, 5);
        addEdge(tree, 5, 9);
        addEdge(tree, 9, 10);
        addEdge(tree, 9, 11);
        addEdge(tree, 11, 12);
        addEdge(tree, 3, 7);
        addEdge(tree, 7, 6);
        addEdge(tree, 7, 8);
 
        int NodesWithFruits[]
            = { 2, 4, 5, 7, 8, 9, 11, 12 };
 
        // Number of nodes with fruits
        int m = NodesWithFruits.length;
 
        int ans = maxFruit(tree, NodesWithFruits, n, m, k);
 
        System.out.println(ans);
    }
}
// This code is contributed by karandeep1234


Output

7

Complexity Analysis:

  • Time Complexity: O(n*n*k) (worst case, eg: 2 level tree with the root having n-1 child nodes)
  • Auxiliary Space: O(n)

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