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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 : 17 Aug, 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 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;
}


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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