Count the nodes of a tree whose weighted string is an anagram of the given string

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weighted string is an anagram with the given string str.

Examples:

Input:

str = “geek”
Output: 2
Only the weighted strings of the nodes 2 and 6
are anagrams of the given string “geek”.



Approach: Perform dfs on the tree and for every node, check if it’s weighted string is anagram with the given string or not, If not then increment the count.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
string s;
int cnt = 0;
  
vector<int> graph[100];
vector<string> weight(100);
  
// Function that return true if both
// the strings are anagram of each other
bool anagram(string x, string s)
{
    sort(x.begin(), x.end());
    sort(s.begin(), s.end());
    if (x == s)
        return true;
    else
        return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If current node's weighted
    // string is an anagram of
    // the given string s
    if (anagram(weight[node], s))
        cnt += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    s = "geek";
  
    // Weights of the nodes
    weight[1] = "eeggk";
    weight[2] = "geek";
    weight[3] = "gekrt";
    weight[4] = "tree";
    weight[5] = "eetr";
    weight[6] = "egek";
  
    // 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 << cnt;
  
    return 0;
}

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Output:

2


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