This "Hard" problem is a more general version of making an array increasing. Here, the condition is that for a given k, the sequence arr[i], arr[i+k], arr[i+2k]... must be non-decreasing for every i from 0 to k-1. You can replace any element with any other positive integer. The goal is to find the minimum number of replacements to satisfy this for all k independent sequences.
Amazon uses this problem to test a candidate's ability to recognize the Longest Non-Decreasing Subsequence (LNDS) pattern. It requires understanding that to minimize replacements in a sequence, you must maximize the number of elements you keep. The number of elements you can keep is exactly the length of the longest non-decreasing subsequence of that sequence.
The algorithmic pattern is a reduction to the Longest Increasing Subsequence (LIS/LNDS) problem. The array is split into k separate subsequences based on the indices (e.g., indices 0, k, 2k..., then 1, k+1, 2k+1...). For each subsequence, you find the length of its Longest Non-Decreasing Subsequence using the O(N log N) Binary Search method. The operations needed for a subsequence is length_of_subsequence - length_of_LNDS. Sum these up for the final answer.
Array: [4, 1, 5, 2, 6, 2], k = 2.
[4, 5, 6]. LNDS length = 3. Ops = 3 - 3 = 0.[1, 2, 2]. LNDS length = 3. Ops = 3 - 3 = 0.
Total operations = 0.
(If k=3 and subsequence was [5, 1, 3], LNDS is [1, 3] length 2, Ops = 3-2 = 1).The most common mistake is attempting to solve this with a simple greedy approach (like the "Easy" version). Because you can replace an element with any value, it's not just about incrementing. Candidates also often use the O(N^2) DP for LIS, which is too slow. Another error is miscounting the indices when extracting the k subsequences.
Master the O(N log N) binary search solution for LIS and LNDS. This is a recurring theme in "Hard" array problems where you need to optimize the number of elements kept. Also, practice the pattern of deconstructing a single array into multiple interleaved sequences, as this appears in many "k-spaced" problems.