Problem Description

You are given an array of integers nums. You can perform the following operation any number of times:

• Remove a strictly increasing subsequence from the array

Your goal is to find the minimum number of operations needed to make the array empty.

A strictly increasing subsequence means selecting elements from the array (maintaining their relative order) where each element is strictly greater than the previous one. For example, if nums = [5, 3, 4, 1, 2], you could remove [3, 4] as one operation since 3 < 4, or remove [1, 2] since 1 < 2.

The challenge is to strategically choose which subsequences to remove so that the total number of operations is minimized. Each operation removes one complete strictly increasing subsequence, and you need to determine the fewest number of such operations to remove all elements from the array.

For instance, if nums = [5, 3, 4, 1, 2], one way would be:

• Operation 1: Remove [3, 4] → remaining array: [5, 1, 2]
• Operation 2: Remove [1, 2] → remaining array: [5]
• Operation 3: Remove [5] → array becomes empty

This takes 3 operations total, which happens to be the minimum for this example.

Intuition

Let's think about this problem from a different angle. Instead of focusing on which subsequences to remove, let's consider what happens when we can't add an element to any existing subsequence.

When we traverse the array from left to right, for each element x, we want to append it to one of the subsequences we're currently building. To maintain the strictly increasing property, we can only append x to a subsequence whose last element is less than x.

Here's the key insight: if we greedily choose to append x to the subsequence with the largest last element that is still less than x, we maximize our chances of being able to add future elements to this subsequence. This is because we're keeping the last element as small as possible while still being valid.

Now, what happens to the last elements of all our subsequences as we build them? If we always make the greedy choice above, the last elements of our subsequences will be in decreasing order. Why? Because when a new element x comes in:

• If x is larger than all existing last elements, we start a new subsequence with x as its last element
• If x can extend some subsequences, we choose the one with the largest valid last element and replace it with x

This decreasing order property is crucial because it allows us to use binary search to quickly find which subsequence to extend.

The minimum number of operations equals the number of subsequences we need to build. Each subsequence corresponds to one removal operation. By greedily minimizing the number of subsequences (through smart placement of each element), we minimize the total number of operations needed.

Think of it like stacking items where each stack must be strictly increasing from bottom to top. We want to minimize the number of stacks. The greedy strategy ensures we use the minimum number of stacks possible.

Learn more about Binary Search patterns.

Solution Approach

Based on our intuition, we need to maintain the last elements of our subsequences in decreasing order. We'll use an array g to store these last elements.

Here's how the algorithm works:

1. Initialize an empty array g: This array will store the last element of each subsequence we're building. The array g will maintain a decreasing order throughout the process.

2. Process each element x in nums:

   • We need to find the position where x should be placed in g
   • Since g is in decreasing order, we want to find the first element in g that is smaller than x
   • This is where binary search comes in

3. Binary Search Implementation:

   l, r = 0, len(g)
   while l < r:
       mid = (l + r) >> 1
       if g[mid] < x:
           r = mid
       else:
           l = mid + 1

   • We search for the leftmost position where g[mid] < x
   • The condition g[mid] < x means we can append x to the subsequence ending with g[mid]
   • After the binary search, l points to the position where we should place x

4. Update the array g:

   if l == len(g):
       g.append(x)
   else:
       g[l] = x

   • If l == len(g), it means x is smaller than all elements in g, so we need to start a new subsequence (append x to g)
   • Otherwise, we replace g[l] with x, which means we're extending the subsequence that previously ended with g[l]

5. Return the result: The length of g represents the minimum number of subsequences needed, which equals the minimum number of operations.

Example walkthrough with nums = [5, 3, 4, 1, 2]:

• Process 5: g = [5]
• Process 3: 3 < 5, append → g = [5, 3]
• Process 4: Find first element < 4 (which is 3 at index 1), replace → g = [5, 4]
• Process 1: 1 is smaller than all, append → g = [5, 4, 1]
• Process 2: Find first element < 2 (which is 1 at index 2), replace → g = [5, 4, 2]
• Result: len(g) = 3

The time complexity is O(n log n) where n is the length of the input array, due to the binary search for each element. The space complexity is O(n) in the worst case when all elements form a decreasing sequence.

Example Walkthrough

Let's walk through the solution with nums = [4, 6, 3, 5, 2]:

Initial state: g = [] (empty array to track last elements of subsequences)

Step 1: Process element 4

• g is empty, so append 4
• g = [4]
• This represents starting our first subsequence ending with 4

Step 2: Process element 6

• Binary search in g = [4] for first element < 6
• Found 4 at index 0, so replace g[0] with 6
• g = [6]
• We extended the subsequence [4] to [4, 6]

Step 3: Process element 3

• Binary search in g = [6] for first element < 3
• No element in g is < 3, so l = len(g) = 1
• Append 3 to g
• g = [6, 3]
• Started a second subsequence with just [3]

Step 4: Process element 5

• Binary search in g = [6, 3] for first element < 5
• Found 3 at index 1, so replace g[1] with 5
• g = [6, 5]
• Extended the second subsequence from [3] to [3, 5]

Step 5: Process element 2

• Binary search in g = [6, 5] for first element < 2
• No element in g is < 2, so l = len(g) = 2
• Append 2 to g
• g = [6, 5, 2]
• Started a third subsequence with just [2]

Final result: len(g) = 3, meaning we need 3 operations

The three subsequences we built were:

• Subsequence 1: [4, 6]
• Subsequence 2: [3, 5]
• Subsequence 3: [2]

Each subsequence represents one removal operation, giving us the minimum of 3 operations needed to empty the array.

Solution Implementation

Time and Space Complexity

The time complexity of this algorithm is O(n log n), where n is the length of the array nums.

The algorithm iterates through each element in nums exactly once, which takes O(n) time. For each element, it performs a binary search on the array g to find the appropriate position. The binary search operation takes O(log k) time, where k is the current size of array g. In the worst case, g can grow up to size n, so the binary search can take up to O(log n) time. Therefore, the overall time complexity is O(n) × O(log n) = O(n log n).

The space complexity is O(n), where n is the length of the array nums.

The algorithm uses an auxiliary array g to store elements. In the worst case, when all elements in nums form a strictly decreasing sequence, the array g will store all n elements from the input array. Therefore, the space complexity is O(n).

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

Common Pitfalls

1. Confusion About Binary Search Direction

The most common pitfall is misunderstanding which comparison operator to use in the binary search. Many developers mistakenly write:

Incorrect Implementation:

if non_increasing_subsequences[mid] > current_num:  # WRONG!
    right = mid
else:
    left = mid + 1

Why this is wrong: We're maintaining subsequences where we can append strictly increasing elements. We need to find where current_num can extend an existing subsequence, which means finding an element that is less than current_num (so current_num can be appended after it).

Correct Implementation:

if non_increasing_subsequences[mid] < current_num:  # CORRECT
    right = mid
else:
    left = mid + 1

2. Misunderstanding the Array's Purpose

Another pitfall is thinking the array non_increasing_subsequences stores actual subsequences or that it must be strictly decreasing.

What the array actually represents:

• Each position i in the array represents a subsequence
• The value at position i is the last element of that subsequence
• The array maintains these last elements in non-increasing order (not strictly decreasing)

For example, with nums = [5, 3, 3, 2]:

• After processing: g = [5, 3, 3, 2]
• This represents 4 different subsequences, each containing one element
• The array can have duplicates (non-increasing, not strictly decreasing)

3. Off-by-One Error in Binary Search Bounds

Incorrect:

left, right = 0, len(non_increasing_subsequences) - 1  # WRONG!

Correct:

left, right = 0, len(non_increasing_subsequences)  # CORRECT

The right bound should be len(array) (not len(array) - 1) because we might need to append to the end of the array when no suitable position is found.

4. Using Wrong Binary Search Template

Some might use a different binary search template that doesn't work correctly for this problem:

Incorrect Template:

while left <= right:  # WRONG for this problem
    mid = (left + right) // 2
    if non_increasing_subsequences[mid] < current_num:
        right = mid - 1
    else:
        left = mid + 1

Correct Template:

while left < right:  # CORRECT
    mid = (left + right) >> 1
    if non_increasing_subsequences[mid] < current_num:
        right = mid
    else:
        left = mid + 1

The correct template ensures we find the leftmost position where we can place current_num, maintaining the non-increasing property of the array.