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.



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

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 <bits/stdc++.h>
using namespace std;
int cnt = 0;
vector<int> graph[100];
vector<string> 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++)
    if (mp.size() == x.size())
        return true;
        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)
        dfs(to, node);
// Driver code
int main()
    // Weights of the nodes
    weight[1] = "abc";
    weight[2] = "aba";
    weight[3] = "bcb";
    weight[4] = "moh";
    weight[5] = "aa";
    // Edges of the tree
    dfs(1, 1);
    cout << cnt;
    return 0;




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