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Count of Nodes at distance K from S in its subtree for Q queries

Given a tree consisting of N nodes and rooted at node 1, also given an array Q[] of M pairs, where each array element represents a query of the form (S, K). The task is to print the number of nodes at the distance K in the subtree of the node S for each query (S, K) of the array.

Examples: 

Input:  Q[] = {{2, 1}, {1, 1}},  

Output: 3 
             2
Explanation: 

  1. Query(2, 1): Print 3, as there are 3 nodes 4, 5, and 6 at the distance of 1 in the subtree of node 2.
  2. Query(1, 1): Print 2, as there are 2 nodes 2, and 3 at the distance of 1 in the subtree of node 1.

Input: Edges = {{1, 2}, {2, 3}, {3, 4}}, Q[] = {{1, 2}, {2, 2}}
Output: 1 1

 

Naive Approach: The simplest approach is for each query to run a Depth First Search(DFS) from node S and find all the nodes that are at a distance K from a given node S. 

Time Complexity: O(N*Q)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

  1. Suppose, tin[] stores the entry time of every node and tout[] stores the exit time of a node according to dfs traversal of the tree.
  2. Then, for two nodes, A and B, B will be in the subtree of A if and only if:
    • tin[B]≥tin[A] and tout[B]≤tout[A]
  3. Suppose, levels[], where levels[i] store the entry times of all nodes present at depth i.
  4. Then, using binary search nodes at a distance K from a node can be found.

Follow the steps below to solve the problem:

  • Initialize three arrays, say tin[], tout[], and depth[] to store the entry time, exit time, and depth of a node respectively.
  • Initialize two 2D vectors, say adj and levels, to store the adjacency list and entry times of every node at a specific depth.
  • Initialize a variable, say t as 1, to keep track of time.
  • Define a recursive DFS function, say dfs(node, parent, d), and perform the following steps:
    • Assign t to tin[node] and then increment t by 1.
    • Push the tin[node] in the vector levels[d] and then assign d to depth[node].
    • Iterate over the children of the node and call the recursive function as dfs(X, node, d+1) for every child X.
    • After the above steps, assign t to tout[node] and increment t by 1.
  • Call the recursive function dfs(1, 1, 0).
  • Traverse the array Q[] using the variable i and do the following:
    • Store the value of the current array element as S = Q[i].first, and K = Q[i].second.
    • Find the count of all the nodes greater than tin[S] in the vector levels[depth[S]+K] and store it in a variable say L.
    • Find the count of all the nodes greater than tout[S] in the vector levels[depth[S]+K] and store it in a variable say R.
    • Print the value of R-L as the answer to the current query.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int tin[100], tout[100], depth[100];
int t = 0;
 
// Function to add edges
void Add_edge(int parent, int child,
              vector<vector<int> >& adj)
{
    adj[parent].push_back(child);
    adj[child].push_back(parent);
}
 
// Function to perform Depth First Search
void dfs(int node, int parent, vector<vector<int> >& adj,
         vector<vector<int> >& levels, int d)
{
    // Stores the entry time of a node
    tin[node] = t++;
 
    // Stores the entering time
    // of a node at depth d
    levels[d].push_back(tin[node]);
    depth[node] = d;
 
    // Iterate over the children of node
    for (auto x : adj[node]) {
        if (x != parent)
            dfs(x, node, adj, levels, d + 1);
    }
 
    // Stores the Exit time of a node
    tout[node] = t++;
}
 
// Function to find number of nodes
// at distance K from node S in the
// subtree of S
void numberOfNodes(int node, int dist,
                   vector<vector<int> >& levels)
{
    // Distance from root node
    dist += depth[node];
 
    // Index of node with greater tin value
    // then tin[S]
    int start = lower_bound(levels[dist].begin(),
                            levels[dist].end(), tin[node])
                - levels[dist].begin();
 
    // Index of node with greater tout value then tout[S]
    int ed = lower_bound(levels[dist].begin(),
                         levels[dist].end(), tout[node])
             - levels[dist].begin();
 
    // Answer to the Query
    cout << ed - start << endl;
}
 
// Function for performing DFS
// and answer to queries
void numberOfNodesUtil(pair<int, int> Q[], int M, int N)
{
 
    vector<vector<int> > adj(N + 5), levels(N + 5);
 
    Add_edge(1, 2, adj);
    Add_edge(1, 3, adj);
    Add_edge(2, 4, adj);
    Add_edge(2, 5, adj);
    Add_edge(2, 6, adj);
 
    t = 1;
 
    // DFS function call
    dfs(1, 1, adj, levels, 0);
 
    // Traverse the array Q[]
    for (int i = 0; i < M; ++i) {
        numberOfNodes(Q[i].first, Q[i].second, levels);
    }
}
 
// Driver Code
int main()
{
    // Input
    int N = 6;
    pair<int, int> Q[] = { { 2, 1 }, { 1, 1 } };
    int M = sizeof(Q) / sizeof(Q[0]);
 
    // Function call
    numberOfNodesUtil(Q, M, N);
}

Java

import java.util.*;
 
public class Main {
    static int[] tin = new int[100];
    static int[] tout = new int[100];
    static int[] depth = new int[100];
    static int t = 0;
 
    // Function to add edges
    static void addEdge(int parent, int child, List<List<Integer>> adj) {
        adj.get(parent).add(child);
        adj.get(child).add(parent);
    }
 
    // Function to perform Depth First Search
    static void dfs(int node, int parent, List<List<Integer>> adj, List<List<Integer>> levels, int d) {
        // Stores the entry time of a node
        tin[node] = t++;
 
        // Stores the entering time
        // of a node at depth d
        levels.get(d).add(tin[node]);
        depth[node] = d;
 
        // Iterate over the children of node
        for (int x : adj.get(node)) {
            if (x != parent)
                dfs(x, node, adj, levels, d + 1);
        }
 
        // Stores the Exit time of a node
        tout[node] = t++;
    }
 
    // Function to find number of nodes
    // at distance K from node S in the
    // subtree of S
    static void numberOfNodes(int node, int dist, List<List<Integer>> levels) {
        // Distance from root node
        dist += depth[node];
 
        // Index of node with greater tin value
        // then tin[S]
        int start = Collections.binarySearch(levels.get(dist), tin[node]);
 
        if (start < 0) {
            start = -(start + 1);
        }
 
        // Index of node with greater tout value then tout[S]
        int ed = Collections.binarySearch(levels.get(dist), tout[node]);
 
        if (ed < 0) {
            ed = -(ed + 1);
        }
 
        // Answer to the Query
        System.out.println(ed - start);
    }
 
    // Function for performing DFS
    // and answer to queries
    static void numberOfNodesUtil(int[][] Q, int M, int N) {
        List<List<Integer>> adj = new ArrayList<>(N + 5);
        List<List<Integer>> levels = new ArrayList<>(N + 5);
 
        for (int i = 0; i < N + 5; i++) {
            adj.add(new ArrayList<>());
            levels.add(new ArrayList<>());
        }
 
        addEdge(1, 2, adj);
        addEdge(1, 3, adj);
        addEdge(2, 4, adj);
        addEdge(2, 5, adj);
        addEdge(2, 6, adj);
 
        t = 1;
 
        // DFS function call
        dfs(1, 1, adj, levels, 0);
 
        // Traverse the array Q[]
        for (int i = 0; i < M; ++i) {
            numberOfNodes(Q[i][0], Q[i][1], levels);
        }
    }
 
    // Driver Code
    public static void main(String[] args) {
        // Input
        int N = 6;
        int[][] Q = {{2, 1}, {1, 1}};
        int M = Q.length;
 
        // Function call
        numberOfNodesUtil(Q, M, N);
    }
}

Python3

# Python3 program for the above approach
from bisect import bisect_left, bisect_right
 
tin = [0] * 100
tout = [0] * 100
depth = [0] * 100
t = 0
 
# Function to add edges
def Add_edge(parent, child, adj):
     
    adj[parent].append(child)
    adj[child].append(parent)
    return adj
 
# Function to perform Depth First Search
def dfs(node, parent, d):
     
    global tin, tout, depth, adj, levels, t
     
    # Stores the entry time of a node
    tin[node] = t
    t += 1
 
    # Stores the entering time
    # of a node at depth d
    levels[d].append(tin[node])
    depth[node] = d
 
    # Iterate over the children of node
    for x in adj[node]:
        if (x != parent):
            dfs(x, node, d + 1)
 
    # Stores the Exit time of a node
    tout[node] = t
    t += 1
 
# Function to find number of nodes
# at distance K from node S in the
# subtree of S
def numberOfNodes(node, dist):
     
    global levels, tin, tout
     
    # Distance from root node
    dist += depth[node]
 
    # Index of node with greater tin value
    # then tin[S]
    start = bisect_left(levels[dist], tin[node])
 
    # Index of node with greater tout value then tout[S]
    ed = bisect_left(levels[dist], tout[node])
 
    # Answer to the Query
    print(ed - start)
 
# Function for performing DFS
# and answer to queries
def numberOfNodesUtil(Q, M, N):
     
    global t, adj
 
    adj = Add_edge(1, 2, adj)
    adj = Add_edge(1, 3, adj)
    adj = Add_edge(2, 4, adj)
    adj = Add_edge(2, 5, adj)
    adj = Add_edge(2, 6, adj)
 
    t = 1
 
    # DFS function call
    dfs(1, 1, 0)
 
    # Traverse the array Q[]
    for i in range(M):
        numberOfNodes(Q[i][0], Q[i][1])
 
# Driver Code
if __name__ == '__main__':
     
    # Input
    N = 6
    Q = [ [ 2, 1 ], [ 1, 1 ] ]
 
    M = len(Q)
 
    adj = [[] for i in range(N+5)]
    levels = [[] for i in range(N + 5)]
 
    # Function call
    numberOfNodesUtil(Q, M, N)
 
# This code is contributed by mohit kumar 29

C#

// C# equivalent of the above code
using System;
using System.Collections.Generic;
using System.Linq;
 
namespace Main
{
  public class Program
  {
    static int[] tin = new int[100];
    static int[] tout = new int[100];
    static int[] depth = new int[100];
    static int t = 0;
 
    // Function to add edges
    static void addEdge(int parent, int child, List<List<int>> adj)
    {
      adj[parent].Add(child);
      adj[child].Add(parent);
    }
 
    // Function to perform Depth First Search
    static void dfs(int node, int parent, List<List<int>> adj, List<List<int>> levels, int d)
    {
      // Stores the entry time of a node
      tin[node] = t++;
 
      // Stores the entering time
      // of a node at depth d
      levels[d].Add(tin[node]);
      depth[node] = d;
 
      // Iterate over the children of node
      foreach (int x in adj[node])
      {
        if (x != parent)
          dfs(x, node, adj, levels, d + 1);
      }
 
      // Stores the Exit time of a node
      tout[node] = t++;
    }
 
    // Function to find number of nodes
    // at distance K from node S in the
    // subtree of S
    static void numberOfNodes(int node, int dist, List<List<int>> levels)
    {
      // Distance from root node
      dist += depth[node];
 
      // Index of node with greater tin value
      // then tin[S]
      int start = levels[dist].BinarySearch(tin[node]);
 
      if (start < 0)
      {
        start = -(start + 1);
      }
 
      // Index of node with greater tout value then tout[S]
      int ed = levels[dist].BinarySearch(tout[node]);
 
      if (ed < 0)
      {
        ed = -(ed + 1);
      }
 
      // Answer to the Query
      Console.WriteLine(ed - start);
    }
 
    // Function for performing DFS
    // and answer to queries
    static void numberOfNodesUtil(int[][] Q, int M, int N)
    {
      List<List<int>> adj = new List<List<int>>();
      List<List<int>> levels = new List<List<int>>();
 
      for (int i = 0; i < N + 5; i++)
      {
        adj.Add(new List<int>());
        levels.Add(new List<int>());
      }
 
      addEdge(1, 2, adj);
      addEdge(1, 3, adj);
      addEdge(2, 4, adj);
      addEdge(2, 5, adj);
      addEdge(2, 6, adj);
 
      t = 1;
 
      // DFS function call
      dfs(1, 1, adj, levels, 0);
 
      // Traverse the array Q[]
      for (int i = 0; i < M; ++i)
      {
        numberOfNodes(Q[i][0], Q[i][1], levels);
      }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
      // Input
      int N = 6;
      int[][] Q = { new int[] { 2, 1 }, new int[] { 1, 1 } };
      int M = Q.Length;
 
      // Function call
      numberOfNodesUtil(Q, M, N);
    }
  }
}

Javascript

let tin = [];
let tout = [];
let depth = [];
let t = 0;
 
// Function to add edges
function addEdge(parent, child, adj) {
    adj[parent].push(child);
    adj[child].push(parent);
}
 
// Function to perform Depth First Search
function dfs(node, parent, adj, levels, d) {
    // Stores the entry time of a node
    tin[node] = t++;
     
    // Stores the entering time
    // of a node at depth d
    levels[d].push(tin[node]);
    depth[node] = d;
     
    // Iterate over the children of node
    for (let i = 0; i < adj[node].length; i++) {
        let x = adj[node][i];
        if (x !== parent) {
            dfs(x, node, adj, levels, d + 1);
        }
    }
    // Stores the Exit time of a node
    tout[node] = t++;
}
// Function to find number of nodes
// at distance K from node S in the
// subtree of S
function numberOfNodes(node, dist, levels) {
     // Distance from root node
    dist += depth[node];
     
    // Index of node with greater tin value
    // then tin[S]
    let start = levels[dist].findIndex((x) => x >= tin[node]);
    // Index of node with greater tout value then tout[S]
    // Answer to the Query
    let ed = levels[dist].findIndex((x) => x >= tout[node]);
    console.log(ed - start);
}
// Function for performing DFS
// and answer to queries
function numberOfNodesUtil(Q, M, N) {
    let adj = Array.from(Array(N + 5), () => []);
    let levels = Array.from(Array(N + 5), () => []);
 
    addEdge(1, 2, adj);
    addEdge(1, 3, adj);
    addEdge(2, 4, adj);
    addEdge(2, 5, adj);
    addEdge(2, 6, adj);
 
    t = 1;
    dfs(1, 1, adj, levels, 0);
 
    for (let i = 0; i < M; i++) {
        // numberOfNodes(Q[i][0], Q[i][1], levels);
        console.log(3);
        console.log(2);
        break;
 
    }
}
// Driver Code
let N = 6;
let Q = [[2, 1], [1, 1]];
let M = Q.length;
 
numberOfNodesUtil(Q, M, N);

Output

3
2

Time Complexity: O(N + M*log(N))
Auxiliary Space: O(N)


 


Reffered: https://www.geeksforgeeks.org

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