Problem Description

Given an array of integers nums, you are allowed to perform the following operation any number of times:

• Remove a strictly increasing subsequence from the array.

Your task is to find the minimum number of operations required to make the array empty.

Intuition

To solve this problem, we need to determine the minimum number of operations required to remove strictly increasing subsequences from the array until it becomes empty.

The key observation is that to perform the least number of operations, each operation should remove the longest possible strictly increasing subsequence, as this would reduce the array's size more significantly with fewer operations.

The approach involves:

1. Sequentially Iterating through the Array: For each element in the array, check where it fits within the context of forming the longest increasing subsequences known so far.

2. Using a Greedy Strategy with Binary Search: Maintain a sequence of potential end elements for various lengths of increasing subsequences. Each time a new number is considered, we should determine where it can extend or modify the existing sequences. For efficient lookup and modification, a binary search is utilized.

3. Building the Solution: If the current number is smaller than any sequence's end, it signifies the start of a new potential subsequence, or it could replace an existing element in an ongoing subsequence to maintain or extend it. Thus, meticulously altering or appending elements efficiently ensures that we eventually find the total number of distinct increasing subsequences possible. This count directly correlates to the minimal number of operations required to empty the array.

This greedy approach capitalizes on maintaining well-defined subsequences, ensuring optimal solution derivation by leveraging the power of binary search for rapid calculation adjustments.

Learn more about Binary Search patterns.

Solution Approach

The problem is approached using a combination of greedy strategy and binary search, which helps efficiently manage and determine the sequences required to minimize the operations.

Step-by-Step Implementation:

1. Initialization:

   • Begin with an empty list g that will store the end elements of the active sequences of increasing subsequences.

2. Iterating through the Array:

   • For each element x in the array nums, determine its position with respect to the sequences maintained in g.

3. Binary Search for Insertion Point:

   • Use a binary search algorithm to find the appropriate position of x in the list g. This position indicates where x can replace an existing end element or be appended:
     • If x can replace an element in g, it suggests that x can extend or form a new potential sequence at that position (favorable and leads to convergence towards longer subsequences).
     • If x cannot replace any element and is larger than all, append x to g, signifying it starts a new sequence.

4. Appending and Replacing:

   • If l, which is the result of the binary search, equals the length of g, append x to g as it extends the longest sequence so far.
   • Otherwise, replace g[l] with x to maintain the minimum possible end element for the sequence of that length, which helps in future sequence formation.

5. Result Calculation:

   • After all elements in nums are processed, the length of g will equal the minimal number of strictly increasing subsequences needed. Consequently, it represents the minimum operations to empty the array.

Here is the algorithm implemented as described:

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        g = []
        for x in nums:
            l, r = 0, len(g)
            while l < r:
                mid = (l + r) >> 1
                if g[mid] < x:
                    r = mid
                else:
                    l = mid + 1
            if l == len(g):
                g.append(x)
            else:
                g[l] = x
        return len(g)

This efficient combination of greedy approach and binary search ensures that the solution is derived with optimal consideration of space and time complexities.

Example Walkthrough

Let's walk through an example to demonstrate the solution approach:

Consider the array nums = [4, 3, 6, 2, 8, 5, 7].

Step 1: Initialization

• Start with an empty list g = [] to track the end elements of potential increasing subsequences.

Step 2: Iterate through the Array

• Element 4:

  • Use binary search on g which is currently empty, so directly append 4 to g.
  • g becomes [4].

• Element 3:

  • Binary search on g = [4]. Since 3 is less than 4, replace 4 with 3 for a new shortest subsequence.
  • g becomes [3].

• Element 6:

  • Binary search on g = [3]. 6 is greater than 3, so append 6 to g.
  • g becomes [3, 6].

• Element 2:

  • Binary search on g = [3, 6]. 2 is less than both, replace 3 with 2.
  • g becomes [2, 6].

• Element 8:

  • Binary search on g = [2, 6]. 8 is greater than 6, so append 8 to g.
  • g becomes [2, 6, 8].

• Element 5:

  • Binary search on g = [2, 6, 8]. 5 is less than 6, replace 6 with 5.
  • g becomes [2, 5, 8].

• Element 7:

  • Binary search on g = [2, 5, 8]. 7 is less than 8, replace 8 with 7.
  • g becomes [2, 5, 7].

Step 3: Result Calculation

• The length of g is 3, indicating that 3 operations are needed to remove all strictly increasing subsequences and empty the array.

This process efficiently segments the array into the minimal number of strictly increasing subsequences using a combination of greedy strategy and binary search.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n log n), where n is the length of the array nums. This is because each element is processed, and the binary search within the current list g takes O(log n) time. For each element, you might need to search the entire g, hence n iterations, leading to the overall time complexity of O(n log n).

The space complexity is O(n), since in the worst case, all elements of nums may end up in g, making the space usage proportional to n.

Learn more about how to find time and space complexity quickly.