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++

`// 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; ` `} ` |

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## Python3

`# Python3 implementation of the approach ` `graph ` `=` `[[] ` `for` `i ` `in` `range` `(` `100` `)] ` `weight ` `=` `[` `0` `] ` `*` `100` ` ` `# Function that returns if the ` `# string x is a pangram ` `def` `Pangram(x): ` ` ` `mp ` `=` `{} ` ` ` `n ` `=` `len` `(x) ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `if` `x[i] ` `not` `in` `mp: ` ` ` `mp[x[i]] ` `=` `0` ` ` `mp[x[i]] ` `+` `=` `1` ` ` `if` `(` `len` `(mp)` `=` `=` `26` `): ` ` ` `return` `True` ` ` `else` `: ` ` ` `return` `False` ` ` `# Function to return the count of nodes ` `# which make pangram with the ` `# sub-tree nodes ` `def` `countTotalPangram(n): ` ` ` `cnt ` `=` `0` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `): ` ` ` `if` `(Pangram(weight[i])): ` ` ` `cnt ` `+` `=` `1` ` ` `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 ` `def` `dfs(node, parent): ` ` ` `for` `to ` `in` `graph[node]: ` ` ` `if` `(to ` `=` `=` `parent): ` ` ` `continue` ` ` `dfs(to, node) ` ` ` `weight[node] ` `+` `=` `weight[to] ` ` ` `# Driver code ` `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` `].append(` `2` `) ` `graph[` `2` `].append(` `3` `) ` `graph[` `2` `].append(` `4` `) ` `graph[` `1` `].append(` `5` `) ` `graph[` `5` `].append(` `6` `) ` ` ` `dfs(` `1` `, ` `1` `) ` `print` `(countTotalPangram(n)) ` ` ` `# This code is contributed by SHUBHAMSINGH10 ` |

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**Output:**

1

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