# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `int` `ans = 0; ` ` `  `vector<``int``> graph; ` `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 = 4; ` `    ``weight = 5; ` `    ``weight = 3; ` `    ``weight = 25; ` `    ``weight = 16; ` `    ``weight = 30; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.push_back(5); ` `    ``graph.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[] 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 `

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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