生成长度为 K 的最长回文子串的长度为 N 的字符串

给定两个整数N和K ( K ≤ N )，任务是获得长度为N的字符串，使得该字符串的回文子串的最大长度为K 。

例子：

Input: N = 5, K = 3 
Output: “abacd” 
Explanation: Palindromic substrings are “a”, “b”, “c”, “d” and “aba”. Therefore, the longest palindromic substring from the given string is of length 3.

Input: N = 8, K = 4 
Output: “abbacdef” 
Explanation: Palindromic subtsrings are “a”, “b”, “c”, “d”, “e”, “f”, “bb”, “abba”. Therefore, the longest palindromic substring from the given string is of length 4.

方法：这个想法基于以下观察，即由单个字符组成的任何长度的字符串总是回文的，例如 {‘a’, ‘bbbbb’, ‘ccc’}。因此，为了生成具有所需条件的字符串，打印 ‘a’ K次，使其具有长度为K的最长回文子串，用非回文序列填充剩余的N – K个插槽。

请按照以下步骤解决问题：

• 打印 ‘a’ 正好K次。
• 考虑一个非回文序列，比如“bcd”。
• 打印字符串。

下面是上述方法的实现：

C++

// C++ program to implement the above approach
 
#include 
using namespace std;
 
// Function to generate a string of
// length N having longest palindromic
// substring of length K
void string_palindrome(int N, int K)
{
 
    // Fill first K characters with 'a'
    for (int i = 0; i < K; i++)
        cout << "a";
 
    // Stores a non-palindromic sequence
    // to be repeated for N - k slots
    string s = "bcd";
 
    // Print N - k remaining characters
    for (int i = 0; i < N - K; i++)
        cout << s[i % 3];
}
 
// Driver Code
int main()
{
 
    // Given N and K
    int N = 5, K = 3;
    string_palindrome(N, K);
 
    return 0;
}

Java

// Java program to implement the above approach
import java.util.*;
 
class GFG
{
 
// Function to generate a String of
// length N having longest palindromic
// subString of length K
static void String_palindrome(int N, int K)
{
 
    // Fill first K characters with 'a'
    for (int i = 0; i < K; i++)
        System.out.print("a");
 
    // Stores a non-palindromic sequence
    // to be repeated for N - k slots
    String s = "bcd";
 
    // Print N - k remaining characters
    for (int i = 0; i < N - K; i++)
        System.out.print(s.charAt(i % 3));
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given N and K
    int N = 5, K = 3;
    String_palindrome(N, K);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement the above approach
 
# Function to generate a string of
# length N having longest palindromic
# substring of length K
def string_palindrome(N, K):
 
    # Fill first K characters with 'a'
    for i in range(K):
        print("a", end = "")
 
    # Stores a non-palindromic sequence
    # to be repeated for N - k slots
    s = "bcd"
 
    # PrN - k remaining characters
    for i in range(N - K):
        print(s[i % 3], end = "")
 
# Driver Code
if __name__ == '__main__':
   
    # Given N and K
    N, K = 5, 3
    string_palindrome(N, K)
 
    # This code is contributed by mohit kumar 29

C#

// C# program to implement the above approach
using System;
class GFG
{
     
    // Function to generate a String of
    // length N having longest palindromic
    // subString of length K
    static void String_palindrome(int N, int K)
    {
     
        // Fill first K characters with 'a'
        for (int i = 0; i < K; i++)
            Console.Write("a");
     
        // Stores a non-palindromic sequence
        // to be repeated for N - k slots
        string s = "bcd";
     
        // Print N - k remaining characters
        for (int i = 0; i < N - K; i++)
            Console.Write(s[i % 3]);
    }
     
    // Driver Code
    public static void Main(string[] args)
    {
     
        // Given N and K
        int N = 5, K = 3;
        String_palindrome(N, K);
    }
}
 
// This code is contributed by AnkThon

Javascript

 
// JavaScript program for above approach
 
    // Function to generate a String of
    // length N having longest palindromic
    // subString of length K
    function String_palindrome(N, K)
    {
      
        // Fill first K characters with 'a'
        for (let i = 0; i < K; i++)
            document.write("a");
      
        // Stores a non-palindromic sequence
        // to be repeated for N - k slots
        let s = "bcd";
      
        // Print N - k remaining characters
        for (let i = 0; i < N - K; i++)
            document.write(s[i % 3]);
    }
 
 
// Driver Code
 
     // Given N and K
        let N = 5, K = 3;
        String_palindrome(N, K);
         

输出：

aaabc

时间复杂度： O(N)
辅助空间： O(1)

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