LeetCode in Kotlin

2111. Minimum Operations to Make the Array K-Increasing

Hard

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 <= 105
• 1 <= arr[i], k <= arr.length

Solution

class Solution {
    fun kIncreasing(a: IntArray, k: Int): Int {
        val n = a.size
        var res = 0
        for (s in 0 until k) {
            val dp: MutableList<Int> = ArrayList()
            var i = s
            while (i < n) {
                if (!bsearch(dp, a[i])) {
                    dp.add(a[i])
                }
                i += k
            }
            res += dp.size
        }
        return n - res
    }

    private fun bsearch(dp: MutableList<Int>, target: Int): Boolean {
        if (dp.isEmpty()) {
            return false
        }
        var lo = 0
        var hi = dp.size - 1
        while (lo < hi) {
            val mid = lo + (hi - lo) / 2
            if (dp[mid] <= target) {
                lo = mid + 1
            } else {
                hi = mid
            }
        }
        if (dp[lo] > target) {
            dp[lo] = target
            return true
        }
        return false
    }
}

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