Count the nodes whose weight is a perfect square

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:

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// 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;
}

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Output:

3


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