Minimum Operations to Make the Array K-Increasing

Problem

You are given a 0-indexed array arr consisting of n positive integers, and a positive integer k.

The array arr is called K-increasing if arr[i-k] <= arr[i] holds for every index i, where k <= i <= n-1.

For example, arr = [4, 1, 5, 2, 6, 2] is K-increasing for k = 2 because:

arr[0] <= arr[2] (4 <= 5)
arr[1] <= arr[3] (1 <= 2)
arr[2] <= arr[4] (5 <= 6)
arr[3] <= arr[5] (2 <= 2)
However, the same arr is not K-increasing for k = 1 (because arr[0] > arr[1]) or k = 3 (because arr[0] > arr[3]).

In one operation, you can choose an index i and change arr[i] into any positive integer.

Return **the *minimum number of operations* required to make the array K-increasing for the given **k.

Example 1:

Input: arr = [5,4,3,2,1], k = 1
Output: 4
Explanation:
For k = 1, the resultant array has to be non-decreasing.
Some of the K-increasing arrays that can be formed are [5,6,7,8,9], [1,1,1,1,1], [2,2,3,4,4]. All of them require 4 operations.
It is suboptimal to change the array to, for example, [6,7,8,9,10] because it would take 5 operations.
It can be shown that we cannot make the array K-increasing in less than 4 operations.

Example 2:

Input: arr = [4,1,5,2,6,2], k = 2
Output: 0
Explanation:
This is the same example as the one in the problem description.
Here, for every index i where 2 <= i <= 5, arr[i-2] <= arr[i].
Since the given array is already K-increasing, we do not need to perform any operations.

Example 3:

Input: arr = [4,1,5,2,6,2], k = 3
Output: 2
Explanation:
Indices 3 and 5 are the only ones not satisfying arr[i-3] <= arr[i] for 3 <= i <= 5.
One of the ways we can make the array K-increasing is by changing arr[3] to 4 and arr[5] to 5.
The array will now be [4,1,5,4,6,5].
Note that there can be other ways to make the array K-increasing, but none of them require less than 2 operations.

Constraints:

1 <= arr.length <= 10^5
1 <= arr[i], k <= arr.length

Solution

class Solution {
    public int kIncreasing(int[] a, int k) {
        int n = a.length;
        int res = 0;
        for (int s = 0; s < k; s++) {
            List<Integer> dp = new ArrayList<>();
            for (int i = s; i < n; i += k) {
                if (!bsearch(dp, a[i])) {
                    dp.add(a[i]);
                }
            }
            res += dp.size();
        }
        return n - res;
    }

    private boolean bsearch(List<Integer> dp, int target) {
        if (dp.isEmpty()) {
            return false;
        }
        int lo = 0;
        int hi = dp.size() - 1;
        while (lo < hi) {
            int mid = lo + (hi - lo) / 2;
            if (dp.get(mid) <= target) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }

        if (dp.get(lo) > target) {
            dp.set(lo, target);
            return true;
        }
        return false;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).