# Sort an array of strings based on count of distinct characters

Given a string array arr[] as input, the task is to print the words sorted by number of distinct characters that occur in the word, followed by length of word.

Note:

• If two words have same number of distinct characters, the word with more total characters comes first.
• If two words have same number of distinct characters and same length, the word that occurs earlier in the sentence must be printed first.

Examples:

Input: arr[] = {“Bananas”, “do”, “not”, “grow”, “in”, “Mississippi”}
Output: do in not Mississippi Bananas grow
Explanation:
After sorting by the number of unique characters and the length the output will be, do in not Mississippi Bananas grow.

Input: arr[] = {“thank”, “you”, “geeks”, “world”}
Output: you geeks thank world
Explanation:
After sorting by the number of unique characters and the length the output will be, you geeks thank world.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use Sorting

• Initialize a map data structure to count all the possible distinct characters from each string of the given array.
• Then sort the array by passing the comparator function, where sorting is done by the number of unique character in word and length of word.
• After sorting is done, print the strings of the array.

For example s = Bananas do not grow in Mississippi”

Word                   Number of unique character           Length of Word

do                                                     2                                2

in                                                     2                                2

not                                                  3                                3

Bananas                                         4                               7

grow                                               4                              4

Mississippi                                    4                              11

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return no of ` `// unique character in a word ` `int` `countDistinct(string s) ` `{ ` `    ``// Initialize map ` `    ``unordered_map<``char``, ``int``> m; ` ` `  `    ``for` `(``int` `i = 0; i < s.length(); i++) { ` `        ``// Count distinct characters ` `        ``m[s[i]]++; ` `    ``} ` ` `  `    ``return` `m.size(); ` `} ` ` `  `// Function to perform sorting ` `bool` `compare(string& s1, string& s2) ` `{ ` `    ``if` `(countDistinct(s1) == countDistinct(s2)) { ` `        ``// Check if size of string 1 ` `        ``// is same as string 2 then ` `        ``// return false because s1 should ` `        ``// not be placed before s2 ` `        ``if` `(s1.size() == s2.size()) { ` `            ``return` `false``; ` `        ``} ` `        ``return` `s1.size() > s2.size(); ` `    ``} ` `    ``return` `countDistinct(s1) < countDistinct(s2); ` `} ` ` `  `// Function to print the sorted array of string ` `void` `printArraystring(string str[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << str[i] << ``" "``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string arr[] = { ``"Bananas"``, ``"do"``, ` `                     ``"not"``, ``"grow"``, ``"in"``, ` `                     ``"Mississippi"` `}; ` `    ``int` `n = ``sizeof``(arr) ` `            ``/ ``sizeof``(arr); ` ` `  `    ``// Function call ` `    ``sort(arr, arr + n, compare); ` ` `  `    ``// Print result ` `    ``printArraystring(arr, n); ` ` `  `    ``return` `0; ` `} `

## C#

 `// C# program of the above approach  ` `using` `System; ` `using` `System.Collections;  ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG{ ` `     `  `// Function to return no of ` `// unique character in a word ` `static` `int` `countDistinct(``string` `s) ` `{ ` ` `  `    ``// Initialize map ` `    ``Dictionary<``char``,  ` `               ``int``> m = ``new` `Dictionary<``char``,  ` `                                       ``int``>(); ` ` `  `    ``for``(``int` `i = 0; i < s.Length; i++)  ` `    ``{ ` `         `  `        ``// Count distinct characters ` `        ``if` `(m.ContainsKey(s[i])) ` `        ``{ ` `            ``m[s[i]]++;  ` `        ``} ` `        ``else` `        ``{ ` `            ``m[s[i]] = 1; ` `        ``} ` `    ``} ` `    ``return` `m.Count; ` `} ` ` `  `static` `int` `compare(``string` `s1, ``string` `s2) ` `{ ` `    ``if` `(countDistinct(s1) == countDistinct(s2))  ` `    ``{ ` `         `  `        ``// Check if size of string 1 ` `        ``// is same as string 2 then ` `        ``// return false because s1 should ` `        ``// not be placed before s2 ` `        ``return` `s2.Length - s1.Length; ` `    ``} ` `    ``else` `    ``{ ` `        ``return` `(countDistinct(s1) -  ` `                ``countDistinct(s2)); ` `    ``} ` `} ` ` `  `// Function to print the sorted array of string ` `static` `void` `printArraystring(``string` `[]str, ``int` `n) ` `{ ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``Console.Write(str[i] + ``" "``); ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``string` `[]arr = { ``"Bananas"``, ``"do"``, ` `                     ``"not"``, ``"grow"``, ` `                     ``"in"``, ``"Mississippi"` `}; ` `    ``int` `n = arr.Length; ` `     `  `    ``// Function call ` `    ``Array.Sort(arr, compare); ` `     `  `    ``// Print result ` `    ``printArraystring(arr, n); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

Output:

```do in not Mississippi Bananas grow
```

Time Complexity: O(n * log n)

Auxiliary Space: O(n)

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