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

• Difficulty Level : Easy
• Last Updated : 18 Dec, 2020

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.

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;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to return no of``// unique character in a word``static` `int` `countDistinct(String s)``{``    ` `    ``// Initialize map``    ``Map m = ``new` `HashMap<>();` `    ``for``(``int` `i = ``0``; i < s.length(); i++)``    ``{``        ` `        ``// Count distinct characters``        ``if` `(m.containsKey(s.charAt(i)))``        ``{``            ``m.put(s.charAt(i),``            ``m.get(s.charAt(i)) + ``1``);``        ``}``        ``else``        ``{``            ``m.put(s.charAt(i), ``1``);``        ``}``    ``}``    ``return` `m.size();``}` `// Function to print the sorted``// array of string``static` `void` `printArraystring(String[] str,``                             ``int` `n)``{``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``System.out.print(str[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String[] arr = { ``"Bananas"``, ``"do"``,``                     ``"not"``, ``"grow"``,``                     ``"in"``, ``"Mississippi"` `};``    ``int` `n = arr.length;` `    ``// Function call``    ``Arrays.sort(arr, ``new` `Comparator()``    ``{``        ``public` `int` `compare(String a, String b)``        ``{``            ``if` `(countDistinct(a) ==``                ``countDistinct(b))``            ``{``                ` `                ``// Check if size of string 1``                ``// is same as string 2 then``                ``// return false because s1 should``                ``// not be placed before s2``                ``return` `(b.length() - a.length());``            ``}``            ``else``            ``{``                ``return` `(countDistinct(a) -``                        ``countDistinct(b));``            ``}``        ``}``    ``});` `    ``// Print result``    ``printArraystring(arr, n);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program of the above approach``import` `functools` `# Function to return no of``# unique character in a word``def` `countDistinct(s):` `    ``# Initialize dictionary``    ``m ``=` `{}` `    ``for` `i ``in` `range``(``len``(s)):` `        ``# Count distinct characters``        ``if` `s[i] ``not` `in` `m:``            ``m[s[i]] ``=` `1``        ``else``:``            ``m[s[i]] ``+``=` `1` `    ``return` `len``(m)` `# Function to perform sorting``def` `compare(a, b):``    ` `    ``if` `(countDistinct(a) ``=``=` `countDistinct(b)):``        ` `        ``# Check if size of string 1``        ``# is same as string 2 then``        ``# return false because s1 should``        ``# not be placed before s2``        ``return` `(``len``(b) ``-` `len``(a))``    ``else``:``        ``return` `(countDistinct(a) ``-` `countDistinct(b))` `# Driver Code``arr ``=` `[ ``"Bananas"``, ``"do"``, ``"not"``,``        ``"grow"``, ``"in"``,``"Mississippi"` `]` `n ``=` `len``(arr)` `# Print result``print``(``*``sorted``(``    ``arr, key ``=` `functools.cmp_to_key(compare)), sep ``=` `' '``)` `# This code is contributed by avanitrachhadiya2155`

## 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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