How do you solve Minimum Operations to Make the Array K-Increasing on LeetCode?

Minimum Operations to Make the Array K-Increasing (LeetCode #2111) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with Unknown time complexity and Unknown space complexity. Key insight: We can break the array into k non-overlapping subsequences. For each subsequence, we can calculate the minimum operations needed to make it non-decreasing using a binary search approach to find the longest increasing subsequence (LIS).

What is the time complexity of Minimum Operations to Make the Array K-Increasing?

The optimal solution for Minimum Operations to Make the Array K-Increasing runs in Unknown time and uses Unknown space.

What algorithm or data structure is used to solve Minimum Operations to Make the Array K-Increasing?

Minimum Operations to Make the Array K-Increasing uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Dynamic Programming, Array.

What are the interview tips for Minimum Operations to Make the Array K-Increasing?

Understand the problem constraints clearly. Practice breaking down the problem into smaller parts. Be prepared to explain your thought process during the interview.

What are common mistakes when solving Minimum Operations to Make the Array K-Increasing?

Not breaking the array into k subsequences. Confusing non-decreasing with strictly increasing.

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

Tags: Array, Binary Search

Related patterns: Binary Search, Dynamic Programming, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks all possible subsequences formed by the given k. It counts how many changes are needed to make each subsequence non-decreasing. This is straightforward but inefficient.

The time complexity is O(n²) because we check each element against its predecessor in the subsequence, leading to potentially n checks for each of the n elements.

Step 1: For each index i from k to n-1, check if arr[i-k] <= arr[i].
Step 2: If the condition fails, increment a counter for operations needed.
Step 3: Return the total count of operations after checking all relevant indices.

For arr = [5, 4, 3, 2, 1] and k = 1: 1. i = 1: arr[0] (5) > arr[1] (4) → operations = 1 2. i = 2: arr[1] (4) > arr[2] (3) → operations = 2 3. i = 3: arr[2] (3) > arr[3] (2) → operations = 3 4. i = 4: arr[3] (2) > arr[4] (1) → operations = 4 Final operations = 4.

Optimal Solution approach

Time complexity: Unknown. Space complexity: Unknown.

We can break the array into k non-overlapping subsequences. For each subsequence, we can calculate the minimum operations needed to make it non-decreasing using a binary search approach to find the longest increasing subsequence (LIS).

Step 1: Create k subsequences by iterating through the array with a step of k.
Step 2: For each subsequence, calculate the length of the longest non-decreasing subsequence (LIS).
Step 3: The minimum operations for each subsequence is the length of the subsequence minus the length of the LIS.
Step 4: Sum the operations for all subsequences and return the total.

Common Mistakes

Not breaking the array into k subsequences.
Confusing non-decreasing with strictly increasing.

Interview Tips

Understand the problem constraints clearly.
Practice breaking down the problem into smaller parts.
Be prepared to explain your thought process during the interview.

#2111

Minimum Operations to Make the Array K-Increasing

Hard

Array↗Binary Search↗Binary Search↗Dynamic Programming↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	Unknown
Space	O(1)	Unknown

💡

Intuition

Time UnknownSpace Unknown

⚙️

Algorithm

4 steps

1Step 1: Create k subsequences by iterating through the array with a step of k.
2Step 2: For each subsequence, calculate the length of the longest non-decreasing subsequence (LIS).
3Step 3: The minimum operations for each subsequence is the length of the subsequence minus the length of the LIS.
4Step 4: Sum the operations for all subsequences and return the total.

solution.py16 lines

1def min_operations(arr, k):
2    from bisect import bisect_right
3    def lis_length(seq):
4        lis = []
5        for x in seq:
6            pos = bisect_right(lis, x)
7            if pos == len(lis):
8                lis.append(x)
9            else:
10                lis[pos] = x
11        return len(lis)
12    operations = 0
13    for start in range(k):
14        subsequence = arr[start::k]
15        operations += len(subsequence) - lis_length(subsequence)
16    return operations

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.