Count N-length arrays made from first M natural numbers whose subarrays can be made palindromic by replacing less than half of its elements
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).
Input: N = 2, M = 3
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
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:
- Initialize an integer variable, say ans = 1.
- Traverse from i = 0 to N – 1 and update ans as ans = ans * (M-i)
- Print ans as the answer.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(1)
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