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: We use the standard boundary-finding template to find the first position where g[mid] < x:

   left, right = 0, len(g) - 1
   first_true_index = -1
   
   while left <= right:
       mid = (left + right) // 2
       if g[mid] < x:  # feasible condition: element is smaller than x
           first_true_index = mid
           right = mid - 1  # search left for earlier position
       else:
           left = mid + 1

   • The feasible condition g[mid] < x identifies positions where we can append x
   • We search for the first (leftmost) such position by moving right = mid - 1 when feasible
   • first_true_index tracks the answer; if no valid position exists, it remains -1

4. Update the array g:

   if first_true_index == -1:
       g.append(x)  # x is smaller than or equal to all elements
   else:
       g[first_true_index] = x  # extend the subsequence at this position

   • If first_true_index == -1, no element in g is smaller than x, so we start a new subsequence
   • Otherwise, we replace g[first_true_index] with x, extending that subsequence

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 is false, no element < 3, append → g = [5, 3]
• Process 4: Find first element < 4 (which is 3 at index 1), replace → g = [5, 4]
• Process 1: No element < 1, 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.

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. Using Wrong Binary Search Template Variant

The most common pitfall is using a non-standard binary search template that doesn't properly track the answer. Many developers mistakenly return left directly without tracking first_true_index:

Incorrect Implementation:

while left < right:
    mid = (left + right) >> 1
    if non_increasing_subsequences[mid] < current_num:
        right = mid
    else:
        left = mid + 1
return left  # WRONG! Doesn't handle edge cases properly

Correct Implementation (Template):

left, right = 0, len(non_increasing_subsequences) - 1
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if non_increasing_subsequences[mid] < current_num:
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

# Use first_true_index, which is -1 if no valid position found

2. Confusion About Binary Search Direction

Another pitfall is misunderstanding which comparison operator to use. We need to find the first element that is less than current_num, so we can extend that subsequence:

Why g[mid] < current_num: We're maintaining subsequences where we can append strictly increasing elements. Finding an element less than current_num means current_num can be appended after it.

3. Misunderstanding the Array's Purpose

The array non_increasing_subsequences doesn't store actual subsequences - it stores the last element of each subsequence. The array maintains these last elements in non-increasing order.

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)

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

When using the template, the right bound should be len(array) - 1 (not len(array)), because we use while left <= right:

Correct:

left, right = 0, len(non_increasing_subsequences) - 1
first_true_index = -1

while left <= right:  # Use <= with right = len - 1
    # ...

When first_true_index remains -1, it means no element is smaller than current_num, so we need to append to create a new subsequence.