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Count the nodes whose weight is a perfect square
  • Last Updated : 15 Sep, 2020

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a perfect Square.

Examples:

Input:

Output: 3
Only the weights of nodes 1, 4 and 5 are perfect squares.

Approach: Perform dfs on the tree and for every node, check if it’s weight is a perfect square or not.

Below is the implementation of the above approach:

C++






// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int ans = 0;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function that returns true
// if n is a perfect square
bool isPerfectSquare(int n)
{
    double x = sqrt(n);
    if (floor(x) != ceil(x))
        return false;
    return true;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(weight[node]))
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    int x = 15;
  
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
  
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
    graph[5].push_back(6);
  
    dfs(1, 1);
  
    cout << ans;
  
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
  
class GFG{
   
static int ans = 0;
   
static Vector<Integer>[] graph = new Vector[100]; 
static int[] weight = new int[100];
   
// Function that returns true
// if n is a perfect square
static boolean isPerfectSquare(int n)
{
    double x = Math.sqrt(n);
    if (Math.floor(x) != Math.ceil(x))
        return false;
    return true;
}
   
// Function to perform dfs
static void dfs(int node, int parent)
{
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(weight[node]))
        ans += 1;
   
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
   
// Driver code
public static void main(String[] args)
{
    int x = 15;
    for (int i = 0; i < 100; i++) 
        graph[i] = new Vector<>();
      
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
   
    // Edges of the tree
    graph[1].add(2);
    graph[2].add(3);
    graph[2].add(4);
    graph[1].add(5);
    graph[5].add(6);
   
    dfs(1, 1);
   
    System.out.print(ans); 
}
}
  
// This code is contributed by Rajput-Ji


Python3




# Python3 implementation of the approach
from math import *
ans = 0
  
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function that returns true
# if n is a perfect square
def isPerfectSquare(n):
    x = sqrt(n)
    if (floor(x) != ceil(x)):
        return False
    return True
  
# Function to perform dfs
def dfs(node, parent):
    global ans
      
    # If weight of the current node 
    # is a perfect square
    if (isPerfectSquare(weight[node])):
        ans += 1;
      
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code
  
x = 15
  
# Weights of the node
weight[1] = 4
weight[2] = 5
weight[3] = 3
weight[4] = 25
weight[5] = 16
weight[6] = 30
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
  
dfs(1, 1)
print(ans)
  
# This code is contributed by SHUBHAMSINGH10


C#




// C# program for the above approach
using System; 
using System.Collections; 
using System.Collections.Generic; 
using System.Text; 
  
class GFG{
      
static int ans = 0;
  
static ArrayList[] graph = new ArrayList[100]; 
static int[] weight = new int[100];
  
// Function that returns true
// if n is a perfect square
static bool isPerfectSquare(int n)
{
    double x = Math.Sqrt(n);
      
    if (Math.Floor(x) != Math.Ceiling(x))
        return false;
          
    return true;
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
      
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(weight[node]))
        ans += 1;
  
    foreach(int to in graph[node]) 
    {
        if (to == parent)
            continue;
              
        dfs(to, node);
    }
}
      
// Driver Code
public static void Main(string[] args)
{
    //int x = 15;
    for(int i = 0; i < 100; i++) 
        graph[i] = new ArrayList();
      
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
  
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
    graph[5].Add(6);
  
    dfs(1, 1);
  
    Console.Write(ans); 
}
}
  
// This code is contributed by rutvik_56


Output:

3

Complexity Analysis:

  • Time Complexity: O(N*logV) where V is the maximum weight of a node in the tree.
    In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) for N nodes in the tree. Also, while processing every node, in order to check if the node value is a perfect square or not, the inbuilt sqrt(V), is being called where V is the weight of the node and this function has a complexity of O(log V). Hence for every node, there is an added complexity of O(log V). Therefore, the total time complexity is O(N*logV).
  • Auxiliary Space: O(1).
    Any extra space is not required, so the space complexity is constant.

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