# Generate a String of having N*N distinct non-palindromic Substrings

Given an even integer N, the task is to construct a string such that the total number of distinct substrings of that string which are not a palindrome equals N2.

Examples:

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.

Approach:

It can be observed that, if the first N characters of a string are same, followed by N identical characters different than the first N characters, then the count of distinct non-palindromic substrings will be N2

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.
Similiarly 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

Therefore, to solve the problem, print ‘a’ as the first N characters of the string and ‘b’ as the next N characters of the string.

Below is the implementation of the above approach:

 `// 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 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 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# 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`

Output:
```aaaabbbb
```

Time Complexity: O(N)
Auxiliary Space: O(1)

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