# 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:

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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` |

## 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.