生成具有 N*N 个不同非回文子串的字符串

给定的偶数整数N，任务是构造一个字符串，使得不是回文该字符串的不同子串的总数等于N ^2。

例子：

Input: N = 2 
Output: aabb 
Explanation: 
All the distinct non-palindromic substrings are ab, abb, aab and aabb. 
Therefore, the count of non-palindromic substrings is 4 = 2 ² 
Input: N = 4 
Output: cccczzzz 
Explanation: 
All distinct non-palindromic substrings of the string are cz, czz, czzz, czzzz, ccz, cczz, cczzz, cczzzz, cccz, ccczz, ccczzz, ccczzzz, ccccz, cccczz, cccczzz, cccczzzz. 
The count of non-palindromic substrings is 16. 

方法：
可以观察到的是，如果一个字符串的第一个N个字符是相同的，随后从所述第一N个字符，那么不同的非回文子串的计数将N ²不同的N个相同的字符。

Proof:

N = 3 
str = “aaabbb” 
The string can be split into two substrings of N characters each: “aaa” and “bbb” 
The first character ‘a’ from the first substring forms N distinct non-palindromic substrings “ab”, “abb”, “abbb” with the second substring. 
Similarly, first two characters “aa” forms N distinct non-palindromic substrings “aab”, “aabb”, “aabbb”. 
Similarly, remaining N – 2 characters of the first substring each form N distinct non-palindromic substrings as well. 
Therefore, the total number of distinct non-palindromic substrings is equal to N². 

因此，为了解决该问题，打印“A”作为字符串和“b”作为字符串的下一个N个字符的前N个字符。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to construct a string
// having N*N non-palindromic substrings
void createString(int N)
{
    for (int i = 0; i < N; i++) {
        cout << 'a';
    }
    for (int i = 0; i < N; i++) {
        cout << 'b';
    }
}
 
// Driver Code
int main()
{
    int N = 4;
 
    createString(N);
    return 0;
}

Java

// Java Program to implement
// the above approach
class GFG{
 
// Function to construct a string
// having N*N non-palindromic substrings
static void createString(int N)
{
    for (int i = 0; i < N; i++)
    {
        System.out.print('a');
    }
    for (int i = 0; i < N; i++)
    {
        System.out.print('b');
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 4;
 
    createString(N);
}
}
 
// This code is contributed by shivanisinghss2110

Python3

# Python3 program to implement
# the above approach
 
# Function to construct a string
# having N*N non-palindromic substrings
def createString(N):
 
    for i in range(N):
        print('a', end = '')
    for i in range(N):
        print('b', end = '')
 
# Driver Code
N = 4
 
createString(N)
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to construct a string
// having N*N non-palindromic substrings
static void createString(int N)
{
    for(int i = 0; i < N; i++)
    {
        Console.Write('a');
    }
    for(int i = 0; i < N; i++)
    {
        Console.Write('b');
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 4;
 
    createString(N);
}
}
 
// This code is contributed by Princi Singh

Javascript

// JavaScript program for the above approach
 
// Function to construct a string
// having N*N non-palindromic substrings
function createString(N)
{
    for (let i = 0; i < N; i++)
    {
        document.write('a');
    }
    for (let i = 0; i < N; i++)
    {
        document.write('b');
    }
}
     
// Driver Code
 
        let N = 4;
   
    createString(N);
         

输出：

aaaabbbb

时间复杂度： O(N)
辅助空间： O(1)

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