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 squarebool isPerfectSquare(int n){ double x = sqrt(n); if (floor(x) != ceil(x)) return false; return true;}// Function to perform dfsvoid 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 codeint 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 approachimport 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 squarestatic boolean isPerfectSquare(int n){ double x = Math.sqrt(n); if (Math.floor(x) != Math.ceil(x)) return false; return true;} // Function to perform dfsstatic 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 codepublic 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 approachfrom math import *ans = 0graph = [[] for i in range(100)]weight = [0] * 100# Function that returns true# if n is a perfect squaredef isPerfectSquare(n): x = sqrt(n) if (floor(x) != ceil(x)): return False return True# Function to perform dfsdef 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 codex = 15# Weights of the nodeweight[1] = 4weight[2] = 5weight[3] = 3weight[4] = 25weight[5] = 16weight[6] = 30# Edges of the treegraph[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 approachusing 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 squarestatic bool isPerfectSquare(int n){ double x = Math.Sqrt(n); if (Math.Floor(x) != Math.Ceiling(x)) return false; return true;}// Function to perform dfsstatic 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 Codepublic 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 |
Javascript
<script>// Javascript implementation of the approach let ans=0; let graph = new Array(100); let weight = new Array(100); for(let i=0;i<100;i++) { graph[i]=[]; weight[i]=0; } // Function that returns true // if n is a perfect square function isPerfectSquare(n) { let x = Math.sqrt(n); if (Math.floor(x) != Math.ceil(x)) return false; return true; } // Function to perform dfs function dfs(node,parent) { // If weight of the current node // is a perfect square if (isPerfectSquare(weight[node])) ans += 1; for(let to=0;to<graph[node].length;to++) { if(graph[node][to] == parent) continue dfs(graph[node][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].push(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); graph[5].push(6); dfs(1, 1); document.write( ans); // This code is contributed by unknown2108 </script> |
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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