计算数组中不同字符小于等于 M 的字符串数

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，找到字符串的不同字符，如果不同字符的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

Javascript

// 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

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)

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