2111. Minimum Operations to Make the Array K-Increasing

Description

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:
    <ul>
	<li><code>arr[0] &lt;= arr[2] (4 &lt;= 5)</code></li>
	<li><code>arr[1] &lt;= arr[3] (1 &lt;= 2)</code></li>
	<li><code>arr[2] &lt;= arr[4] (5 &lt;= 6)</code></li>
	<li><code>arr[3] &lt;= arr[5] (2 &lt;= 2)</code></li>
</ul>
</li>
<li>However, the same <code>arr</code> is not K-increasing for <code>k = 1</code> (because <code>arr[0] &gt; arr[1]</code>) or <code>k = 3</code> (because <code>arr[0] &gt; arr[3]</code>).</li>

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

Solutions

Solution 1

Python Code
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class Solution:
    def kIncreasing(self, arr: List[int], k: int) -> int:
        def lis(arr):
            t = []
            for x in arr:
                idx = bisect_right(t, x)
                if idx == len(t):
                    t.append(x)
                else:
                    t[idx] = x
            return len(arr) - len(t)

        return sum(lis(arr[i::k]) for i in range(k))

Java Code
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class Solution {
    public int kIncreasing(int[] arr, int k) {
        int n = arr.length;
        int ans = 0;
        for (int i = 0; i < k; ++i) {
            List<Integer> t = new ArrayList<>();
            for (int j = i; j < n; j += k) {
                t.add(arr[j]);
            }
            ans += lis(t);
        }
        return ans;
    }

    private int lis(List<Integer> arr) {
        List<Integer> t = new ArrayList<>();
        for (int x : arr) {
            int idx = searchRight(t, x);
            if (idx == t.size()) {
                t.add(x);
            } else {
                t.set(idx, x);
            }
        }
        return arr.size() - t.size();
    }

    private int searchRight(List<Integer> arr, int x) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr.get(mid) > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

C++ Code
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class Solution {
public:
    int kIncreasing(vector<int>& arr, int k) {
        int ans = 0, n = arr.size();
        for (int i = 0; i < k; ++i) {
            vector<int> t;
            for (int j = i; j < n; j += k) t.push_back(arr[j]);
            ans += lis(t);
        }
        return ans;
    }

    int lis(vector<int>& arr) {
        vector<int> t;
        for (int x : arr) {
            auto it = upper_bound(t.begin(), t.end(), x);
            if (it == t.end())
                t.push_back(x);
            else
                *it = x;
        }
        return arr.size() - t.size();
    }
};

Go Code
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func kIncreasing(arr []int, k int) int {
	searchRight := func(arr []int, x int) int {
		left, right := 0, len(arr)
		for left < right {
			mid := (left + right) >> 1
			if arr[mid] > x {
				right = mid
			} else {
				left = mid + 1
			}
		}
		return left
	}

	lis := func(arr []int) int {
		var t []int
		for _, x := range arr {
			idx := searchRight(t, x)
			if idx == len(t) {
				t = append(t, x)
			} else {
				t[idx] = x
			}
		}
		return len(arr) - len(t)
	}

	n := len(arr)
	ans := 0
	for i := 0; i < k; i++ {
		var t []int
		for j := i; j < n; j += k {
			t = append(t, arr[j])
		}
		ans += lis(t)
	}
	return ans
}