# Count the number of strings in an array whose distinct characters are less than equal to M

• Last Updated : 16 Aug, 2022

Given an array of strings arr[] and an integer M, the task is to count the strings whose count of distinct characters is less than M.

Examples:

Input: arr[] = {“ADAM”, “JOHNSON”, “COOL”}, M = 3
Output:
Explanation:
There are two such strings whose count of distinct characters is less than M.
Count of Distinct(“ADAM”) = 3
Count of Distinct(“COOL”) = 3

Input: arr[] = {“HERBIVORES”, “AEROPLANE”, “GEEKSFORGEEKS”}, M = 7
Output:
Explanation:
There are two such strings whose count of distinct characters is less than M.
Count of Distinct(“AEROPLANE”) = 7
Count of Distinct(“GEEKSFORGEEKS”) = 7

Approach: The idea is to iterate over all the strings and find the distinct characters of the string, If the count of the distinct characters in the string is less than or equal to the given value of the M, then increment the count by 1.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to count``// the number of strings in the``// array whose distinct characters``// is less than or equal to M``#include ``#include ``using` `namespace` `std;` `// Function to count the strings``// whose distinct characters``// count is less than M``void` `distinct(string S[], ``int` `M, ``int` `n)``{``    ``int` `count = 0;` `    ``// Loop to iterate over all``    ``// the strings of the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Distinct characters in the``        ``// String with the help of set``        ``set<``char``> set1;``        ``for``(``int` `j = 0; j < S[i].length(); j++)``        ``{``            ``if` `(set1.find(S[i][j]) == set1.end())``                ``set1.insert(S[i][j]);``        ``}``        ``int` `c = set1.size();` `        ``// Checking if its less``        ``// than or equal to M``        ``if` `(c <= M)``            ``count += 1;``    ``}``    ``cout << (count);``}` `// Driver Code``int` `main()``{``    ``string S[] = { ``"HERBIVORES"``, ``"AEROPLANE"``,``                   ``"GEEKSFORGEEKS"` `};``    ``int` `M = 7;``    ``int` `n = ``sizeof``(S) / ``sizeof``(S[0]);``    ` `    ``distinct(S, M, n);` `    ``return` `0;``}` `// This code is contributed by chitranayal`

## Java

 `// Java implementation to count``// the number of strings in the``// array whose distinct characters``// is less than or equal to M``import` `java.util.*;` `class` `GFG{` `// Function to count the strings``// whose distinct characters``// count is less than M``public` `static` `void` `distinct(String[] S, ``int` `M)``{``    ``int` `count = ``0``;``    ` `    ``// Loop to iterate over all``    ``// the strings of the array``    ``for``(``int` `i = ``0``; i < S.length; i++)``    ``{``        ` `    ``// Distinct characters in the``    ``// String with the help of set``    ``Set set = ``new` `HashSet<>();``    ``for``(``int` `j = ``0``; j < S[i].length(); j++)``    ``{``        ``if` `(!set.contains(S[i].charAt(j)))``            ``set.add(S[i].charAt(j));``    ``}``    ``int` `c = set.size();``        ` `    ``// Checking if its less``    ``// than or equal to M``    ``if` `(c <= M)``        ``count += ``1``;``    ``}``    ``System.out.println(count);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String S[] = { ``"HERBIVORES"``, ``"AEROPLANE"``,``                ``"GEEKSFORGEEKS"` `};``    ``int` `M = ``7``;` `    ``distinct(S, M);``}``}` `// This code is contributed by jrishabh99`

## Python3

 `# Python3 implementation to count``# the number of strings in the``# array whose distinct characters``# is less than or equal to M` `# Function to count the strings``# whose distinct characters``# count is less than M``def` `distinct(S, M):``    ``count ``=` `0``    ` `    ``# Loop to iterate over all``    ``# the strings of the array``    ``for` `i ``in` `range` `(``len``(S)):``        ` `        ``# Distinct characters in the``        ``# String with the help of set``        ``c ``=` `len``(``set``([d ``for` `d ``in` `S[i]]))``        ` `        ``# Checking if its less``        ``# than or equal to M``        ``if` `(c <``=` `M):``            ``count ``+``=` `1``    ``print``(count)` `# Driver Code``if` `__name__``=``=` `'__main__'``:``    ` `    ``S ``=` `[``"HERBIVORES"``, ``"AEROPLANE"``,``        ``"GEEKSFORGEEKS"``]``    ``M ``=` `7` `    ``distinct(S, M)`

## C#

 `// C# implementation to count``// the number of strings in the``// array whose distinct characters``// is less than or equal to M``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to count the strings``// whose distinct characters``// count is less than M``public` `static` `void` `distinct(``string``[] S, ``int` `M)``{``    ``int` `count = 0;``    ` `    ``// Loop to iterate over all``    ``// the strings of the array``    ``for``(``int` `i = 0; i < S.Length; i++)``    ``{``        ` `        ``// Distinct characters in the``        ``// String with the help of set``        ``HashSet<``char``> ``set` `= ``new` `HashSet<``char``>();``        ``for``(``int` `j = 0; j < S[i].Length; j++)``        ``{``            ``if` `(!``set``.Contains(S[i][j]))``                ``set``.Add(S[i][j]);``        ``}``        ``int` `c = ``set``.Count;``            ` `        ``// Checking if its less``        ``// than or equal to M``        ``if` `(c <= M)``            ``count += 1;``    ``}``    ``Console.Write(count);``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``string` `[]S = { ``"HERBIVORES"``, ``"AEROPLANE"``,``                   ``"GEEKSFORGEEKS"` `};``    ``int` `M = 7;` `    ``distinct(S, M);``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`2`

Performance Analysis:

• Time Complexity: O(N * M*logM), where M is the maximum length of the string.
• Auxiliary Space: O(M)

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