Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes

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 when concatenated with the strings of the sub-tree nodes becomes a pangram.
Pangram: A pangram is a sentence containing every letter of the English Alphabet.



Output: 1
Only the weighted string of sub-tree of node 1 makes the pangram.

Approach: Perform dfs on the tree and update the weight of every node such that it stores its weight concatenated with the weights of the sub-tree nodes. Then, count the nodes whose updated weighted string forms a pangram.

Below is the implementation of the above approach:





// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
vector<int> graph[100];
vector<string> weight(100);
// Function that returns if the
// string x is a pangram
bool Pangram(string x)
    map<char, int> mp;
    int n = x.size();
    for (int i = 0; i < n; i++)
    if (mp.size() == 26)
        return true;
        return false;
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
int countTotalPangram(int n)
    int cnt = 0;
    for (int i = 1; i <= n; i++)
        if (Pangram(weight[i]))
    return cnt;
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
void dfs(int node, int parent)
    for (int to : graph[node]) {
        if (to == parent)
        dfs(to, node);
        weight[node] += weight[to];
// Driver code
int main()
    int n = 6;
    // Weights of the nodes
    weight[1] = "abcde";
    weight[2] = "fghijkl";
    weight[3] = "abcdefg";
    weight[4] = "mnopqr";
    weight[5] = "stuvwxy";
    weight[6] = "zabcdef";
    // Edges of the tree
    dfs(1, 1);
    cout << countTotalPangram(n);
    return 0;




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