Count of nodes having odd divisors in the given subtree for Q queries

Given a N-ary Tree and Q queries where each query contains a node of the N-ary tree, the task is to count the number of nodes that have an odd number of divisors in the subtree for Q queries. 

Examples: 

Input: 
 

Output: 1 3 0 1 



Explanation: 
Query 1: In the subtree rooted at node 100, there is only one node which is 100 which have 9 divisors {1, 2, 4, 5, 10, 20, 25, 50, 100}. Therefore, there is only one node having the odd number of divisors. 
Query 2: In the subtree rooted at node 4, there are 5 nodes out of which 3 nodes are having an odd number of divisors. That is {4, 9, 100} 
Query 3: In the subtree rooted at node 5, there is only one node which is 5 which has two divisors. Therefore, there zero nodes having an odd number of divisors. 

Naive Approach: A simple solution is to traverse the subtree for each query and find the count of nodes that are having an odd number of divisors.

Efficient Approach: The idea is to pre-compute the count of an odd number of divisors for each subtree and storing the count in hash-map. To pre-compute the count of nodes having an odd number of divisors we can use Depth First Search Traversal. Finally, to check that the current node is having an odd number of divisors or not we can use the fact that every perfect square number has an odd number of divisors.

Below is the implementation of the above approach:

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// C++ implementation to count the
// number of nodes havving odd
// number of divisors for each query
 
#include <bits/stdc++.h>
using namespace std;
 
#define N 100001
 
// Adjacency list
// for tree.
vector<int> adj[N];
 
// Array for values and
// answer at ith node.
int a[N], ans[N];
 
// Function to check whether N
// has odd divisors or not
bool hasOddNumberOfDivisors(int n)
{
    if ((double)sqrt(n) == (int)sqrt(n))
        return true;
    return false;
}
 
// DFS function to pre-compute
// the answers
int dfs(int node, int parent)
{
    // Initialize the count
    int count = 0;
    for (auto i = adj[node].begin(); i != adj[node].end();
         ++i) {
        if (*i != parent) {
 
            // Repeat for every child
            count += dfs(*i, node);
        }
    }
 
    // Increase the count if current node
    // has odd number of divisors
    if (hasOddNumberOfDivisors(a[node]))
        ++count;
 
    ans[node] = count;
    return count;
}
 
// Driver Code
int main()
{
 
    int n = 5, i;
    vector<int> q = { 4, 1, 5, 3 };
 
    // Adjacency List
    adj[1].push_back(2);
    adj[2].push_back(1);
    adj[2].push_back(3);
    adj[3].push_back(2);
    adj[3].push_back(4);
    adj[4].push_back(3);
    adj[1].push_back(5);
    adj[5].push_back(1);
 
    a[1] = 4;
    a[2] = 9;
    a[3] = 14;
    a[4] = 100;
    a[5] = 5;
 
    // Function call
    dfs(1, -1);
 
    for (int i = 0; i < q.size(); i++) {
        cout << ans[q[i]] << " ";
    }
    return 0;
}
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// Java implementation to
// count the number of nodes
// havving odd number of
// divisors for each query
import java.util.*;
class GFG{
 
static final int N = 100001;
 
// Adjacency list
// for tree.
static Vector<Integer> []adj =
              new Vector[N];
 
// Array for values and
// answer at ith node.
static int []a = new int[N];
static int []ans = new int[N];
 
// Function to check whether N
// has odd divisors or not
static boolean hasOddNumberOfDivisors(int n)
{
  if ((double)Math.sqrt(n) ==
      (int)Math.sqrt(n))
    return true;
  return false;
}
 
// DFS function to
// pre-compute the answers
static int dfs(int node,
               int parent)
{
  // Initialize the count
  int count = 0;
  for (int i : adj[node])
  {
    if (i != parent)
    {
      // Repeat for every child
      count += dfs(i, node);
    }
  }
 
  // Increase the count if
  // current node has odd
  // number of divisors
  if (hasOddNumberOfDivisors(a[node]))
    ++count;
 
  ans[node] = count;
  return count;
}
 
// Driver Code
public static void main(String[] args)
{
  int n = 5;
  int[] q = {4, 1, 5, 3};
   
  for (int i = 0; i < adj.length; i++)
    adj[i] = new Vector<Integer>();
   
  // Adjacency List
  adj[1].add(2);
  adj[2].add(1);
  adj[2].add(3);
  adj[3].add(2);
  adj[3].add(4);
  adj[4].add(3);
  adj[1].add(5);
  adj[5].add(1);
 
  a[1] = 4;
  a[2] = 9;
  a[3] = 14;
  a[4] = 100;
  a[5] = 5;
 
  // Function call
  dfs(1, -1);
 
  for (int i = 0; i < q.length; i++)
  {
    System.out.print(ans[q[i]] + " ");
  }
}
}
 
// This code is contributed by Rajput-Ji
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# Python3 implementation to count the
# number of nodes havving odd
# number of divisors for each query
import math
 
N = 100001
  
# Adjacency list
# for tree.
adj = [[] for i in range(N)]
  
# Array for values and
# answer at ith node.
a = [0 for i in range(N)]
 
ans = [0 for i in range(N)]
  
# Function to check whether N
# has odd divisors or not
def hasOddNumberOfDivisors(n):
 
    if (math.sqrt(n) == int(math.sqrt(n))):
        return True
         
    return False
  
# DFS function to pre-compute
# the answers
def dfs(node, parent):
 
    # Initialize the count
    count = 0
     
    for i in adj[node]:
        if (i != parent):
  
            # Repeat for every child
            count += dfs(i, node)
             
    # Increase the count if current node
    # has odd number of divisors
    if (hasOddNumberOfDivisors(a[node])):
        count += 1
  
    ans[node] = count
     
    return count
 
# Driver Code
if __name__=="__main__":
  
    n = 5
    i = 0
     
    q = [ 4, 1, 5, 3 ]
  
    # Adjacency List
    adj[1].append(2)
    adj[2].append(1)
    adj[2].append(3)
    adj[3].append(2)
    adj[3].append(4)
    adj[4].append(3)
    adj[1].append(5)
    adj[5].append(1)
  
    a[1] = 4
    a[2] = 9
    a[3] = 14
    a[4] = 100
    a[5] = 5
     
    # Function call
    dfs(1, -1)
     
    for i in range(len(q)):
        print(ans[q[i]], end = ' ')
  
# This code is contributed by rutvik_56
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// C# implementation to
// count the number of nodes
// havving odd number of
// divisors for each query
using System;
using System.Collections.Generic;
class GFG{
 
static readonly int N = 100001;
 
// Adjacency list
// for tree.
static List<int> []adj =
            new List<int>[N];
 
// Array for values and
// answer at ith node.
static int []a = new int[N];
static int []ans = new int[N];
 
// Function to check whether N
// has odd divisors or not
static bool hasOddNumberOfDivisors(int n)
{
  if ((double)Math.Sqrt(n) ==
      (int)Math.Sqrt(n))
    return true;
  return false;
}
 
// DFS function to
// pre-compute the answers
static int dfs(int node,
               int parent)
{
  // Initialize the count
  int count = 0;
  foreach (int i in adj[node])
  {
    if (i != parent)
    {
      // Repeat for every child
      count += dfs(i, node);
    }
  }
 
  // Increase the count if
  // current node has odd
  // number of divisors
  if (hasOddNumberOfDivisors(a[node]))
    ++count;
 
  ans[node] = count;
  return count;
}
 
// Driver Code
public static void Main(String[] args)
{
  int n = 5;
  int[] q = {4, 1, 5, 3};
   
  for (int i = 0;
           i < adj.Length; i++)
    adj[i] = new List<int>();
   
  // Adjacency List
  adj[1].Add(2);
  adj[2].Add(1);
  adj[2].Add(3);
  adj[3].Add(2);
  adj[3].Add(4);
  adj[4].Add(3);
  adj[1].Add(5);
  adj[5].Add(1);
 
  a[1] = 4;
  a[2] = 9;
  a[3] = 14;
  a[4] = 100;
  a[5] = 5;
 
  // Function call
  dfs(1, -1);
 
  for (int i = 0;
           i < q.Length; i++)
  {
    Console.Write(ans[q[i]] + " ");
  }
}
}
 
// This code is contributed by Rajput-Ji
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Output
1 3 0 1 










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