# 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:

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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:

## C++

`// 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;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` ` ` `static` `String s;` ` ` `static` `int` `cnt = ` `0` `;` ` ` `static` `Vector<Integer>[] graph = ` `new` `Vector[` `100` `];` ` ` `static` `String[] weight = ` `new` `String[` `100` `];` ` ` `// Function that return true if both` ` ` `// the Strings are anagram of each other` ` ` `static` `boolean` `anagram(String x, String s)` ` ` `{` ` ` `x = sort(x);` ` ` `s = sort(s);` ` ` `if` `(x.equals(s))` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` ` ` `}` ` ` `static` `String sort(String inputString)` ` ` `{` ` ` `// convert input string to char array` ` ` `char` `tempArray[] = inputString.toCharArray();` ` ` `// sort tempArray` ` ` `Arrays.sort(tempArray);` ` ` `// return new sorted string` ` ` `return` `new` `String(tempArray);` ` ` `}` ` ` `// Function to perform dfs` ` ` `static` `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` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `s = ` `"geek"` `;` ` ` `for` `(` `int` `i = ` `0` `; i < ` `100` `; i++)` ` ` `graph[i] = ` `new` `Vector<Integer>();` ` ` ` ` `// 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` `].add(` `2` `);` ` ` `graph[` `2` `].add(` `3` `);` ` ` `graph[` `2` `].add(` `4` `);` ` ` `graph[` `1` `].add(` `5` `);` ` ` `graph[` `5` `].add(` `6` `);` ` ` `dfs(` `1` `, ` `1` `);` ` ` `System.out.print(cnt);` ` ` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 implementation of the approach` `cnt ` `=` `0` `graph ` `=` `[[] ` `for` `i ` `in` `range` `(` `100` `)]` `weight ` `=` `[` `0` `] ` `*` `100` `# Function that return true if both` `# the strings are anagram of each other` `def` `anagram(x, s):` ` ` `x ` `=` `sorted` `(` `list` `(x))` ` ` `s ` `=` `sorted` `(` `list` `(s))` ` ` `if` `(x ` `=` `=` `s):` ` ` `return` `True` ` ` `else` `:` ` ` `return` `False` `# Function to perform dfs` `def` `dfs(node, parent):` ` ` `global` `cnt, s` ` ` ` ` `# If weight of the current node` ` ` `# string is an anagram of` ` ` `# the given string s` ` ` `if` `(anagram(weight[node], s)):` ` ` `cnt ` `+` `=` `1` ` ` `for` `to ` `in` `graph[node]:` ` ` `if` `(to ` `=` `=` `parent):` ` ` `continue` ` ` `dfs(to, node)` `# Driver code` `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` `].append(` `2` `)` `graph[` `2` `].append(` `3` `)` `graph[` `2` `].append(` `4` `)` `graph[` `1` `].append(` `5` `)` `graph[` `5` `].append(` `6` `)` `dfs(` `1` `, ` `1` `)` `print` `(cnt)` `# This code is contributed by SHUBHAMSINGH10` |

## C#

`// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG` `{` ` ` `static` `String s;` ` ` `static` `int` `cnt = 0;` ` ` `static` `List<` `int` `>[] graph = ` `new` `List<` `int` `>[100];` ` ` `static` `String[] weight = ` `new` `String[100];` ` ` `// Function that return true if both` ` ` `// the Strings are anagram of each other` ` ` `static` `bool` `anagram(String x, String s)` ` ` `{` ` ` `x = sort(x);` ` ` `s = sort(s);` ` ` `if` `(x.Equals(s))` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` ` ` `}` ` ` `static` `String sort(String inputString)` ` ` `{` ` ` `// convert input string to char array` ` ` `char` `[]tempArray = inputString.ToCharArray();` ` ` `// sort tempArray` ` ` `Array.Sort(tempArray);` ` ` `// return new sorted string` ` ` `return` `new` `String(tempArray);` ` ` `}` ` ` `// Function to perform dfs` ` ` `static` `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;` ` ` `foreach` `(` `int` `to ` `in` `graph[node])` ` ` `{` ` ` `if` `(to == parent)` ` ` `continue` `;` ` ` `dfs(to, node);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `s = ` `"geek"` `;` ` ` `for` `(` `int` `i = 0; i < 100; i++)` ` ` `graph[i] = ` `new` `List<` `int` `>();` ` ` ` ` `// 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].Add(2);` ` ` `graph[2].Add(3);` ` ` `graph[2].Add(4);` ` ` `graph[1].Add(5);` ` ` `graph[5].Add(6);` ` ` `dfs(1, 1);` ` ` `Console.Write(cnt);` ` ` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` ` ` `// Javascript implementation of the approach` `let s;` `let cnt = 0;` `let graph = ` `new` `Array(100);` `let weight = ` `new` `Array(100);` `for` `(let i = 0; i < 100; i++)` `{` ` ` `graph[i] = [];` ` ` `weight[i] = 0;` `}` `const sort1 = str => str.split(` `''` `).sort((a, b) => a.localeCompare(b)).join(` `''` `);` `// Function that return true if both` `// the strings are anagram of each other` `function` `anagram(x, s)` `{` ` ` `x = sort1(x);` ` ` `s = sort1(s);` ` ` ` ` `if` `(x == s)` ` ` `return` `true` `;` ` ` `else` ` ` `return` `false` `;` `}` `// Function to perform dfs` `function` `dfs(node, parent)` `{` ` ` `// If current node's weighted` ` ` `// string is an anagram of` ` ` `// the given string s` ` ` `if` `(anagram(weight[node], s))` ` ` `cnt += 1;` ` ` `for` `(let to = 0; to < graph[node].length; to++)` ` ` `{` ` ` `if` `(graph[node][to] == parent)` ` ` `continue` ` ` `dfs(graph[node][to], node); ` ` ` `}` `}` `// Driver code` ` ` `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(2);` ` ` `graph[2].push(3);` ` ` `graph[2].push(4);` ` ` `graph[1].push(5);` ` ` `graph[5].push(6);` ` ` `dfs(1, 1);` ` ` `document.write(cnt);` ` ` `// This code is contributed by Dharanendra L V.` ` ` `</script>` |

**Output:**

2

__Complexity Analysis:__

**Time Complexity :**O(N*(S*log(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 sort() function is used which has a complexity of O(S*log(S)) where S is the length of the weighted string. Therefore, the time complexity is O(N*(S*log(S))) where S is the maximum length of the weight string in the tree.**Auxiliary Space :**O(1).

Any extra space is not required, so the space complexity is constant.