2111 - Minimum Operations to Make the Array K-Increasing (Hard)

Problem Link

https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/

Problem Statement

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

Approach 1: Longest Increasing Subsequence

We can break input vector into $k$ groups $ai,ai+k,ai+2∗k,...$for each $i<k$. Calculate the LIS (Longest Increasing Subsequence) on each group and compare the length with the target size. We need to perform $a.size()−lengthOfLIS(a)$operations to make it K-increasing.

Written by @wingkwong

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = (int) nums.size();
        vector<int> lis;
        for(int i = 0; i < n; i++) {
            auto it = upper_bound(lis.begin(), lis.end(), nums[i]);
            if(it == lis.end()) lis.push_back(nums[i]);
            else *it = nums[i];
        }
        return (int) lis.size();
    }
    
    int kIncreasing(vector<int>& arr, int k) {
        int ans = 0, n = arr.size();
        for (int i = 0; i < k; i++) {
            vector<int> a;
            for (int j = i; j < n; j += k) {
                a.push_back(arr[j]);
            }
            ans += a.size() - lengthOfLIS(a);
        }
        return ans;
    }
};