# Count the nodes of a tree whose weighted string does not contain any duplicate characters

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights does not contain any duplicate character.

Examples:

Input: Output: 2
Only the strings of the node 1 and 4 contains unique strings.

## 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 contains duplicate char or not, If not then increment the count.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `cnt = 0; ` ` `  `vector<``int``> graph; ` `vector weight(100); ` ` `  `// Function that returns true if the ` `// string contains unique characters ` `bool` `uniqueChars(string x) ` `{ ` `    ``map<``char``, ``int``> mp; ` `    ``int` `n = x.size(); ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``mp[x[i]]++; ` `    ``if` `(mp.size() == x.size()) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Function to perform dfs ` `void` `dfs(``int` `node, ``int` `parent) ` `{ ` `    ``// If weighted string of the current ` `    ``// node contains unique characters ` `    ``if` `(uniqueChars(weight[node])) ` `        ``cnt += 1; ` ` `  `    ``for` `(``int` `to : graph[node]) { ` `        ``if` `(to == parent) ` `            ``continue``; ` `        ``dfs(to, node); ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``// Weights of the nodes ` `    ``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; ` `} `

Output:

```2
```

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