# Count N-length arrays made from first M natural numbers whose subarrays can be made palindromic by replacing less than half of its elements

• Difficulty Level : Medium
• Last Updated : 16 Apr, 2021

Given two integer N and M, the task is to find the count of arrays of size N with elements from the range [1, M] in which all subarrays of length greater than 1 can be made palindromic by replacing less than half of its elements i.e., floor(length/2).

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: N = 2, M = 3
Output: 6
Explanation:
There are 9 arrays possible of length 2 using values 1 to 3 i.e. [1, 1], [1, 2], [1, 3], [2, 1][2, 2], [2, 3], [3, 1], [3, 2], [3, 3].
All of these arrays except [1, 1], [2, 2] and [3, 3] have subarrays of length greater than 1 which requires 1 operation to make them palindrome. So the required answer is 9 – 3 = 6.

Input: N = 5, M = 10
Output: 30240

Approach: The problem can be solved based on the following observations:

• It is possible that the maximum permissible number of operations required to make an array a palindrome is floor(size(array)/2).
• It can be observed that by choosing a subarray, starting and ending with the same value, the number of operations needed to make it a palindrome will be less than floor(size of subarray)/2.
• Therefore, the task is reduced to finding the number of arrays of size N using integer values in the range [1, M], which do not contain any duplicate elements, which can be easily done by finding the permutation of M with N i.e. MpN, which is equal to M * (M – 1) * (M – 2) * … * (M – N + 1).

Follow the steps below to solve the problem:

1. Initialize an integer variable, say ans = 1.
2. Traverse from i = 0 to N – 1 and update ans as ans = ans * (M-i)
3. Print ans as the answer.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;``typedef` `long` `long` `ll;` `// Function to find the number of arrays``// following the given condition``void` `noOfArraysPossible(ll N, ll M)``{``    ``// Initialize answer``    ``ll ans = 1;` `    ``// Calculate nPm``    ``for` `(ll i = 0; i < N; ++i) {``        ``ans = ans * (M - i);``    ``}` `    ``// Print ans``    ``cout << ans;``}` `// Driver Code``int` `main()``{` `    ``// Given N and M``    ``ll N = 2, M = 3;` `    ``// Function Call``    ``noOfArraysPossible(N, M);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG``{` `// Function to find the number of arrays``// following the given condition``static` `void` `noOfArraysPossible(``int` `N, ``int` `M)``{``    ``// Initialize answer``    ``int` `ans = ``1``;` `    ``// Calculate nPm``    ``for` `(``int` `i = ``0``; i < N; ++i)``    ``{``        ``ans = ans * (M - i);``    ``}` `    ``// Print ans``    ``System.out.print(ans);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{` `    ``// Given N and M``    ``int` `N = ``2``, M = ``3``;` `    ``// Function Call``    ``noOfArraysPossible(N, M);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program for the above approach` `# Function to find the number of arrays``# following the given condition``def` `noOfArraysPossible(N, M):``    ` `    ``# Initialize answer``    ``ans ``=` `1`` ` `    ``# Calculate nPm``    ``for` `i ``in` `range``(N):``        ``ans ``=` `ans ``*` `(M ``-` `i)``        ` `    ``# Print ans``    ``print``(ans)`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``# Given N and M``    ``N ``=` `2``    ``M ``=` `3``    ` `    ``# Function Call``    ``noOfArraysPossible(N, M)``    ` `# This code is contributed by jana_sayantan`

## C#

 `// C# program to implement``// the above approach ``using` `System;` `class` `GFG{``     ` `// Function to find the number of arrays``// following the given condition``static` `void` `noOfArraysPossible(``int` `N, ``int` `M)``{``    ``// Initialize answer``    ``int` `ans = 1;`` ` `    ``// Calculate nPm``    ``for` `(``int` `i = 0; i < N; ++i)``    ``{``        ``ans = ans * (M - i);``    ``}`` ` `    ``// Print ans``    ``Console.Write(ans);``}`` ` `// Driver Code``public` `static` `void` `Main()``{``    ``// Given N and M``    ``int` `N = 2, M = 3;`` ` `    ``// Function Call``    ``noOfArraysPossible(N, M);``}``}` `// This code is contributed by susmitakundugoaldanga`

## Javascript

 ``
Output:
`6`

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

My Personal Notes arrow_drop_up