# 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:
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 ` `using` `namespace` `std;`   `vector<``int``> graph[100];` `vector 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;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG{` `    `  `@SuppressWarnings``(``"unchecked"``)` `static` `Vector []graph = ``new` `Vector[``100``];` `static` `String []weight = ``new` `String[``100``];`   `// Function that returns if the` `// String x is a pangram` `static` `boolean` `Pangram(String x)` `{` `    ``HashMap mp = ``new` `HashMap<>();` `    ``int` `n = x.length();`   `    ``for``(``int` `i = ``0` `; i < n; i++)` `    ``{` `        ``if` `(mp.containsKey(x.charAt(i)))` `        ``{` `            ``mp.put(x.charAt(i), ` `            ``mp.get(x.charAt(i)) + ``1``);` `        ``}` `        ``else` `        ``{` `            ``mp.put(x.charAt(i), ``1``);` `        ``}` `    ``}` `    ``if` `(mp.size() == ``26``)` `        ``return` `true``;` `    ``else` `        ``return` `false``;` `}`   `// Function to return the count of nodes` `// which make pangram with the` `// sub-tree nodes` `static` `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` `static` `void` `dfs(``int` `node, ``int` `parent)` `{` `    ``for``(``int` `to : graph[node]) ` `    ``{` `        ``if` `(to == parent)` `            ``continue``;` `            `  `        ``dfs(to, node);` `        ``weight[node] += weight[to];` `    ``}` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``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"``;`   `    ``for``(``int` `i = ``0``; i < graph.length; i++)` `        ``graph[i] = ``new` `Vector();` `        `  `    ``// Edges of the tree` `    ``graph[``1``].add(``2``);` `    ``graph[``2``].add(``3``);` `    ``graph[``2``].add(``4``);` `    ``graph[``1``].add(``5``);` `    ``graph[``5``].add(``6``);`   `    ``dfs(``1``, ``1``);`   `    ``System.out.print(countTotalPangram(n));` `}` `}`   `// This code is contributed by Amit Katiyar`

## 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`

## C#

 `// C# implementation of ` `// the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{    `   `static` `List<``int``> []graph = ` `            ``new` `List<``int``>[100];` `static` `String []weight = ` `                ``new` `String[100];`   `// Function that returns if the` `// String x is a pangram` `static` `bool` `Pangram(String x)` `{` `  ``Dictionary<``char``, ` `             ``int``> mp = ``new` `Dictionary<``char``, ` `                                      ``int``>();` `  ``int` `n = x.Length;`   `  ``for``(``int` `i = 0 ; i < n; i++)` `  ``{` `    ``if` `(mp.ContainsKey(x[i]))` `    ``{` `      ``mp[x[i]] = mp[x[i]] + 1;` `    ``}` `    ``else` `    ``{` `      ``mp.Add(x[i], 1);` `    ``}` `  ``}` `  ``if` `(mp.Count == 26)` `    ``return` `true``;` `  ``else` `    ``return` `false``;` `}`   `// Function to return the ` `// count of nodes which ` `// make pangram with the` `// sub-tree nodes` `static` `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` `static` `void` `dfs(``int` `node, ``int` `parent)` `{` `  ``foreach``(``int` `to ``in` `graph[node]) ` `  ``{` `    ``if` `(to == parent)` `      ``continue``;`   `    ``dfs(to, node);` `    ``weight[node] += weight[to];` `  ``}` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `  ``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"``;`   `  ``for``(``int` `i = 0; ` `          ``i < graph.Length; i++)` `    ``graph[i] = ``new` `List<``int``>();`   `  ``// Edges of the tree` `  ``graph[1].Add(2);` `  ``graph[2].Add(3);` `  ``graph[2].Add(4);` `  ``graph[1].Add(5);` `  ``graph[5].Add(6);`   `  ``dfs(1, 1);` `  ``Console.Write(countTotalPangram(n));` `}` `} `   `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output:

`1`

Complexity Analysis:

• Time Complexity: O(N*S).
In dfs, every node of the tree is processed once, and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the Pangram() function is used for every node which has a complexity of O(S) where S is the sum of the length of all weight strings in a subtree and since this is done for every node, the overall time complexity for this part becomes O(N*S). Therefore, the final time complexity is O(N*S).
• Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.

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