计算数组中不同字符小于等于M的字符串的数量

给定一个字符串数组arr []和一个整数M ，任务是计算不同字符数小于M的字符串。
例子：

Input: arr[] = {“ADAM”, “JOHNSON”, “COOL”}, M = 3 
Output: 2 
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: 2 
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

方法：我们的想法是遍历所有的字符串，找到字符串的明显不同的字符，如果不同的字符字符串中的计数小于或等于M的给定值，然后加1计数。

下面是上述方法的实现：

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 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 set = new HashSet();
        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

输出：

2

性能分析：

• 时间复杂度： O(N * M)，其中M是字符串的最大长度。
• 辅助空间： O(1)