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

  • Last Updated : 25 May, 2021

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:

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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 <bits/stdc++.h>
#include <set>
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<Character> 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




<script>
// Javascript 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
function distinct(S, M, n) {
    let count = 0;
 
    // Loop to iterate over all
    // the strings of the array
    for (let i = 0; i < n; i++) {
 
        // Distinct characters in the
        // String with the help of set
        let set1 = new Set();
        for (let j = 0; j < S[i].length; j++) {
            if (!set1.has(S[i][j]))
                set1.add(S[i][j]);
        }
        let c = set1.size;
 
        // Checking if its less
        // than or equal to M
        if (c <= M)
            count += 1;
    }
    document.write(count);
}
 
// Driver Code
 
let S = ["HERBIVORES", "AEROPLANE",
    "GEEKSFORGEEKS"];
let M = 7;
let n = S.length;
 
distinct(S, M, n);
 
// This code is contributed by _saurabh_jaiswal
</script>
Output: 
2

 

Performance Analysis: 

  • Time Complexity: O(N * M), where M is the maximum length of the string.
  • Auxiliary Space: O(1)



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