产生的大小为N，其大小为M的各子串具有正好k个不同字符的字符串

给定 3 个正整数N 、 M和K 。任务是构造一个由小写字母组成的长度为N的字符串，使得每个长度为M 的子字符串恰好有K 个不同的字母。
例子：

Input: N = 5, M = 2, K = 2 
Output: abade 
Explanation: 
Each substring of size 2 “ab”, “ba”, “ad”, “de” have 2 distinct letters.
Input: N = 7, M = 5, K = 3 
Output: abcaaab 
Explanation: 
Each substring of size 5 “tleel”, “leelt”, “eelte” have 3 distinct letters. 
 

方法：在一个大小为 N 的字符串中，每个大小为 M 的子串必须包含恰好 K 个不同的字母 –

• 构造一个具有 K 个不同字母的字符串，从 ‘a’ 到 ‘z’ 到 M 的大小，并将其余的字母如 ‘a’..
• 因为我们生成了一个大小为 M 的字符串，其中有 K 个不同的值。现在，继续重复它，直到我们达到 N 的字符串大小。

下面是上述方法的实现：

C++

// C++ program to generate a string of size N
// whose each substring of size M
// has exactly K distinct characters
 
#include 
using namespace std;
 
// Function to generate the string
string generateString(int N, int M, int K)
{
 
    // Declare empty string
    string s = "";
 
    // counter for M
    int cnt1 = 0;
 
    // counter for K
    int cnt2 = 0;
 
    // Loop to generate string size of N
    for (int i = 0; i < N; i++) {
        cnt1++;
        cnt2++;
 
        // Generating K distinct
        // letters one by one
        if (cnt1 <= M) {
            if (cnt2 <= K) {
                s = s + char(96 + cnt1);
            }
 
            // After generating b distinct letters,
            // append rest a-b letters as 'a'
            else {
                s = s + 'a';
            }
        }
 
        // Reset the counter value
        // and repeat the process
        else {
            cnt1 = 1;
            cnt2 = 1;
            s = s + 'a';
        }
    }
 
    // return final result string
    return s;
}
 
// Driver code
int main()
{
    int N = 7, M = 5, K = 3;
 
    cout << generateString(N, M, K) << endl;
 
    return 0;
}

Java

// Java program to generate a String of size N
// whose each subString of size M
// has exactly K distinct characters
import java.util.*;
class GFG{
 
// Function to generate the String
static String generateString(int N, int M, int K)
{
 
    // Declare empty String
    String s = "";
 
    // counter for M
    int cnt1 = 0;
 
    // counter for K
    int cnt2 = 0;
 
    // Loop to generate String size of N
    for (int i = 0; i < N; i++)
    {
        cnt1++;
        cnt2++;
 
        // Generating K distinct
        // letters one by one
        if (cnt1 <= M)
        {
            if (cnt2 <= K)
            {
                s = s + (char)(96 + cnt1);
            }
 
            // After generating b distinct letters,
            // append rest a-b letters as 'a'
            else
            {
                s = s + 'a';
            }
        }
 
        // Reset the counter value
        // and repeat the process
        else
        {
            cnt1 = 1;
            cnt2 = 1;
            s = s + 'a';
        }
    }
 
    // return final result String
    return s;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 7, M = 5, K = 3;
 
    System.out.println(generateString(N, M, K));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to generate a string of size N
# whose each substring of size M
# has exactly K distinct characters
# Function to generate the string
def generateString(N, M, K):
 
    # Declare empty string
    s = ""
 
    # counter for M
    cnt1 = 0
 
    # counter for K
    cnt2 = 0
 
    # Loop to generate string size of N
    for i in range (N):
        cnt1 += 1
        cnt2 += 1
 
        # Generating K distinct
        # letters one by one
        if (cnt1 <= M):
            if (cnt2 <= K):
                s = s + chr(96 + cnt1)
             
            # After generating b distinct letters,
            # append rest a-b letters as 'a'
            else:
                s = s + 'a'
 
        # Reset the counter value
        # and repeat the process
        else:
            cnt1 = 1
            cnt2 = 1
            s = s + 'a'
 
    # return final result string
    return s
 
# Driver code
if __name__ == "__main__": 
    N = 7
    M = 5
    K = 3
    print (generateString(N, M, K))
     
# This code is contributed by Chitranayal

C#

// C# program to generate a String of
// size N whose each subString of size
// M has exactly K distinct characters
using System;
 
class GFG{
 
// Function to generate the String
static String generateString(int N, int M, int K)
{
 
    // Declare empty String
    String s = "";
 
    // Counter for M
    int cnt1 = 0;
 
    // Counter for K
    int cnt2 = 0;
 
    // Loop to generate String size of N
    for(int i = 0; i < N; i++)
    {
       cnt1++;
       cnt2++;
        
       // Generating K distinct
       // letters one by one
       if (cnt1 <= M)
       {
           if (cnt2 <= K)
           {
               s = s + (char)(96 + cnt1);
           }
            
           // After generating b distinct letters,
           // append rest a-b letters as 'a'
           else
           {
               s = s + 'a';
           }
       }
        
       // Reset the counter value
       // and repeat the process
       else
       {
           cnt1 = 1;
           cnt2 = 1;
           s = s + 'a';
       }
    }
     
    // Return readonly result String
    return s;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 7, M = 5, K = 3;
 
    Console.WriteLine(generateString(N, M, K));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to generate
// a String of size N
// whose each subString of size M
// has exactly K distinct characters
 
    // Function to generate the String
    function generateString(N,M,K)
    {
        // Declare empty string
    let s = "";
  
    // counter for M
    let cnt1 = 0;
  
    // counter for K
    let cnt2 = 0;
  
    // Loop to generate string size of N
    for (let i = 0; i < N; i++) {
        cnt1++;
        cnt2++;
  
        // Generating K distinct
        // letters one by one
        if (cnt1 <= M) {
            if (cnt2 <= K) {
                s = s + String.fromCharCode(96 + cnt1);
            }
  
            // After generating b distinct letters,
            // append rest a-b letters as 'a'
            else {
                s = s + 'a';
            }
        }
  
        // Reset the counter value
        // and repeat the process
        else {
            cnt1 = 1;
            cnt2 = 1;
            s = s + 'a';
        }
    }
  
    // return final result string
    return s;
    }
     
    // Driver code
    let N = 7, M = 5, K = 3;
    document.write( generateString(N, M, K))
     
     
    // This code is contributed
    // by avanitrachhadiya2155
     

输出：

abcaaab

时间复杂度： O(N)
空间复杂度： O(1)

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