# Count the nodes of the given tree whose weighted string is a palindrome

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights are palindrome.

Examples:

```Input: Output: 3
Only the weights of the nodes 2, 3 and 5 are palindromes.
```

## 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 string is palindrome or not. If yes then increment the count.

Implementation:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `cnt = 0; ` ` `  `vector<``int``> graph; ` `vector weight(100); ` ` `  `// Function that returns true ` `// if x is a palindrome ` `bool` `isPalindrome(string x) ` `{ ` `    ``int` `n = x.size(); ` `    ``for` `(``int` `i = 0; i < n / 2; i++) { ` `        ``if` `(x[i] != x[n - 1 - i]) ` `            ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to perform dfs ` `void` `dfs(``int` `node, ``int` `parent) ` `{ ` ` `  `    ``// Weight of the current node ` `    ``string x = weight[node]; ` ` `  `    ``// If the weight is a palindrome ` `    ``if` `(isPalindrome(x)) ` `        ``cnt += 1; ` ` `  `    ``for` `(``int` `to : graph[node]) { ` `        ``if` `(to == parent) ` `            ``continue``; ` `        ``dfs(to, node); ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``// Weights of the node ` `    ``weight = ``"abc"``; ` `    ``weight = ``"aba"``; ` `    ``weight = ``"bcb"``; ` `    ``weight = ``"moh"``; ` `    ``weight = ``"aa"``; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.push_back(5); ` ` `  `    ``dfs(1, 1); ` ` `  `    ``cout << cnt; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `cnt = ``0``;  ` ` `  `static` `Vector> graph = ``new` `Vector>();  ` `static` `Vector weight = ``new` `Vector();  ` ` `  `// Function that returns true  ` `// if x is a palindrome  ` `static` `boolean` `isPalindrome(String x)  ` `{  ` `    ``int` `n = x.length();  ` `    ``for` `(``int` `i = ``0``; i < n / ``2``; i++) ` `    ``{  ` `        ``if` `(x.charAt(i) != x.charAt(n - ``1` `- i))  ` `            ``return` `false``;  ` `    ``}  ` `    ``return` `true``;  ` `}  ` ` `  `// Function to perform dfs  ` `static` `void` `dfs(``int` `node, ``int` `parent)  ` `{  ` ` `  `    ``// Weight of the current node  ` `    ``String x = weight.get(node);  ` `     `  ` `  `    ``// If the weight is a palindrome  ` `    ``if` `(isPalindrome(x))  ` `        ``cnt += ``1``;  ` ` `  `    ``for` `(``int` `i=``0``;i()); ` `     `  `    ``// Edges of the tree  ` `    ``graph.get(``1``).add(``2``);  ` `    ``graph.get(``2``).add(``3``);  ` `    ``graph.get(``2``).add(``4``);  ` `    ``graph.get(``1``).add(``5``);  ` `    ``dfs(``1``, ``1``);  ` ` `  `    ``System.out.println( cnt);  ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 implementation of the approach ` `cnt ``=` `0` ` `  `graph ``=` `[``0``] ``*` `100` `for` `i ``in` `range``(``100``): ` `    ``graph[i] ``=` `[] ` ` `  `weight ``=` `[``"0"``] ``*` `100` ` `  `# Function that returns true ` `# if x is a palindrome ` `def` `isPalindrome(x): ` `    ``n ``=` `len``(x) ` ` `  `    ``for` `i ``in` `range``(``0``, n ``/``/` `2``): ` `        ``if` `x[i] !``=` `x[n ``-` `1` `-` `i]: ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Function to perform dfs ` `def` `dfs(node, parent): ` `    ``global` `cnt ` ` `  `    ``# Weight of the current node ` `    ``x ``=` `weight[node] ` ` `  `    ``# If the weight is a palindrome ` `    ``if` `(isPalindrome(x)): ` `        ``cnt ``+``=` `1` ` `  `    ``for` `to ``in` `graph[node]: ` `        ``if` `to ``=``=` `parent: ` `            ``continue` `        ``dfs(to, node) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``# Weights of the node ` `    ``weight[``0``] ``=` `"" ` `    ``weight[``1``] ``=` `"abc"` `    ``weight[``2``] ``=` `"aba"` `    ``weight[``3``] ``=` `"bcb"` `    ``weight[``4``] ``=` `"moh"` `    ``weight[``5``] ``=` `"aa"` ` `  `    ``# Edges of the tree ` `    ``graph[``1``].append(``2``) ` `    ``graph[``2``].append(``3``) ` `    ``graph[``2``].append(``4``) ` `    ``graph[``1``].append(``5``) ` ` `  `    ``dfs(``1``, ``1``) ` ` `  `    ``print``(cnt) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` ` `  `    ``static` `int` `cnt = 0;  ` `     `  `    ``static` `List> graph = ``new` `List>();  ` `    ``static` `List weight = ``new` `List();  ` `     `  `    ``// Function that returns true ` `    ``// if x is a palindrome ` `    ``static` `bool` `isPalindrome(``string` `x) ` `    ``{ ` `        ``int` `n = x.Length; ` `        ``for` `(``int` `i = 0; i < n / 2; i++) ` `        ``{ ` `            ``if` `(x[i] != x[n - 1 - i]) ` `                ``return` `false``; ` `        ``} ` `        ``return` `true``; ` `    ``} ` `     `  `    ``// Function to perform dfs  ` `    ``static` `void` `dfs(``int` `node, ``int` `parent)  ` `    ``{  ` `     `  `        ``// Weight of the current node  ` `        ``String x = weight[node];  ` `     `  `        ``// If the weight is a palindrome  ` `        ``if` `(isPalindrome(x))  ` `            ``cnt += 1;  ` `     `  `        ``for` `(``int` `i = 0; i < graph[node].Count; i++)  ` `        ``{  ` `            ``if` `(graph[node][i] == parent)  ` `                ``continue``;  ` `            ``dfs(graph[node][i], node);  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String []args) ` `    ``{  ` `     `  `        ``// Weights of the node  ` `        ``weight.Add( ``""``);  ` `        ``weight.Add( ``"abc"``);  ` `        ``weight.Add( ``"aba"``);  ` `        ``weight.Add( ``"bcb"``);  ` `        ``weight.Add( ``"moh"``);  ` `        ``weight.Add( ``"aa"``);  ` `         `  `        ``for``(``int` `i = 0; i < 100; i++) ` `        ``graph.Add(``new` `List<``int``>()); ` `     `  `        ``// Edges of the tree  ` `        ``graph.Add(2);  ` `        ``graph.Add(3);  ` `        ``graph.Add(4);  ` `        ``graph.Add(5);  ` `     `  `        ``dfs(1, 1);  ` `     `  `        ``Console.WriteLine( cnt);  ` `     `  `    ``}  ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```3
```

Complexity Analysis:

• Time Complexity: O(N*Len) where Len is the maximum length of the weighted string of a node in the given 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, processing of every node involves traversing the weighted string of that node once, thus adding a complexity of O(Len) where Len is the length of the weighted string. Therefore, the total time complexity is O(N*Len).
• Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.

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