# 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.

**Examples:**

Input:

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++) ` ` ` `mp[x[i]]++; ` ` ` `if` `(mp.size() == 26) ` ` ` `return` `true` `; ` ` ` `else` ` ` `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])) ` ` ` `cnt++; ` ` ` `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) ` ` ` `continue` `; ` ` ` `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 ` ` ` `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 << countTotalPangram(n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1

## Recommended Posts:

- Subtree of all nodes in a tree using DFS
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Convert a Binary Tree such that every node stores the sum of all nodes in its right subtree
- Change a Binary Tree so that every node stores sum of all nodes in left subtree
- Count the nodes in the given tree whose weight is even
- Count the nodes in the given tree whose weight is a power of two
- Determine the count of Leaf nodes in an N-ary tree
- Count the nodes in the given tree whose weight is prime
- Count the nodes of the given tree whose weight has X as a factor
- Count the number of nodes at given level in a tree using BFS.
- Count Non-Leaf nodes in a Binary Tree
- Count the number of nodes at a given level in a tree using DFS
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes in the given tree whose weight is even parity
- Count nodes with two children at level L in a Binary Tree

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.