找到每个子串的字符串正好有 K 个不同的字符

给定两个整数N和K 。任务是找到长度为 N的字符串，使得每个长度大于等于K 的子字符串都恰好有K 个不同的字符。
例子：

Input: N=10, K=3
Output : ABCABCABCA
Explanation:
The output string has 3 distinct characters.

Input : N=20, K=7
Output : ABCDEFGABCDEFGABCDEF
Explanation:
The output string has 7 distinct characters.

方法：
为了解决上面提到的问题，主要思想是打印不同的元素直到长度K ，然后重复相同的元素直到N 。
下面是上述方法的实现：

C++

// C++ Program to Find the
// String having each substring
// with exactly K distinct characters
#include 
using namespace std;
 
// Function to find the
// required output string
void findString(int N, int K)
{
    // Each element at index
    // i is modulus of K
    for (int i = 0; i < N; i++) {
 
        cout << char('A' + i % K);
    }
}
 
// Driver code
int main()
{
    // initialise integers N and K
    int N = 10;
    int K = 3;
 
    findString(N, K);
 
    return 0;
}

Java

// Java program to find the
// string having each substring
// with exactly K distinct characters
import java.io.*;
 
class GFG {
 
// Function to find the
// required output string
static void findString(int N, int K)
{
     
    // Each element at index
    // i is modulus of K
    for (int i = 0; i < N; i++)
    {
        System.out.print((char)('A' + i % K));
    }
}
 
// Driver code
public static void main(String[] args)
{
    // Initialise integers N and K
    int N = 10;
    int K = 3;
    findString(N, K);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 Program to Find the
# String having each substring
# with exactly K distinct characters
 
# Function to find the
# required output string
def findString(N, K) :
    # Each element at index
    # i is modulus of K
    for i in range(N) :
 
        print(chr(ord('A') + i % K),end="");
 
# Driver code
if __name__ == "__main__" :
    # initialise integers N and K
    N = 10;
    K = 3;
 
    findString(N, K);
     
# This code is contributed by AnkitRai01

C#

// C# program to find the
// string having each substring
// with exactly K distinct characters
using System;
 
class GFG {
 
// Function to find the
// required output string
static void findString(int N, int K)
{
     
    // Each element at index
    // i is modulus of K
    for(int i = 0; i < N; i++)
    {
       Console.Write((char)('A' + i % K));
    }
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Initialise integers N and K
    int N = 10;
    int K = 3;
 
    findString(N, K);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

    // Javascript Program to Find the
    // String having each substring
    // with exactly K distinct characters
     
    // Function to find the
    // required output string
    function findString(N, K)
    {
     
        // Each element at index
        // i is modulus of K
        for (let i = 0; i < N; i++)
        {
            document.write(String.fromCharCode('A'.charCodeAt() + i % K));
        }
    }
     
    // initialise integers N and K
    let N = 10;
    let K = 3;
   
    findString(N, K);
 
// This code is contributed by divyesh072019.

输出：

ABCABCABCA

时间复杂度： O(N)

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