Problem Description

You are given an integer array nums (0-indexed) and an integer k.

In each operation, you can remove one occurrence of the smallest element from nums.

Your task is to find the minimum number of operations required to make all remaining elements in the array greater than or equal to k.

For example, if nums = [2, 5, 1, 3, 4] and k = 4, the elements less than 4 are [2, 1, 3]. You would need to remove these 3 elements (which takes 3 operations) to ensure all remaining elements are at least 4. The answer would be 3.

The solution works by simply counting how many elements in the array are less than k. Since we always remove the smallest element in each operation, and we want all remaining elements to be ≥ k, we need to remove exactly those elements that are < k. The solution sum(x < k for x in nums) counts all elements less than k, giving us the minimum number of operations needed.

Intuition

The key insight is recognizing that we always remove the smallest element in each operation. This means we're removing elements in ascending order.

Think about it this way: if we want all remaining elements to be ≥ k, then any element that is < k must eventually be removed. It doesn't matter in which order we encounter these elements in the array - they all need to go.

Since we can only remove the smallest element at each step, and we need to remove all elements less than k, the process would look like:

1. Remove the smallest element (if it's < k)
2. Remove the next smallest element (if it's < k)
3. Continue until all elements < k are gone

The crucial realization is that we don't actually need to simulate this removal process. We can directly count how many elements are less than k because:

• Every element < k will need to be removed
• Every element ≥ k will remain in the array
• The order of removal doesn't affect the total count

Therefore, the minimum number of operations equals the count of elements less than k. This leads us to the elegant one-liner solution: sum(x < k for x in nums), which counts all elements where x < k is true.

Solution Approach

The solution uses a simple traversal and counting approach. We iterate through the array once and count the number of elements that are less than k.

The implementation leverages Python's generator expression with the sum() function:

return sum(x < k for x in nums)

Here's how this works step by step:

1. Traversal: We iterate through each element x in the array nums

2. Comparison: For each element x, we evaluate the boolean expression x < k

   • If x < k, the expression evaluates to True (which equals 1 in Python)
   • If x >= k, the expression evaluates to False (which equals 0 in Python)

3. Counting: The sum() function adds up all the boolean values

   • Each True contributes 1 to the sum
   • Each False contributes 0 to the sum
   • The result is the total count of elements less than k

Time Complexity: O(n) where n is the length of the array, as we traverse the array exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for the counting operation (the generator expression doesn't create an intermediate list).

This approach is optimal because we must examine every element at least once to determine if it's less than k, making O(n) the best possible time complexity for this problem.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums = [3, 7, 2, 9, 1, 5] and k = 5

Step 1: Identify elements less than k We need to find all elements that are less than 5:

• 3 < 5 ✓ (needs removal)
• 7 < 5 ✗ (can stay)
• 2 < 5 ✓ (needs removal)
• 9 < 5 ✗ (can stay)
• 1 < 5 ✓ (needs removal)
• 5 < 5 ✗ (can stay, equal to k)

Elements less than 5: [3, 2, 1] → Count = 3

Step 2: Understand the removal process If we were to simulate the actual operations:

• Operation 1: Remove 1 (smallest element) → [3, 7, 2, 9, 5]
• Operation 2: Remove 2 (now smallest) → [3, 7, 9, 5]
• Operation 3: Remove 3 (now smallest) → [7, 9, 5]

After 3 operations, all remaining elements [7, 9, 5] are ≥ 5.

Step 3: Apply the direct counting solution Instead of simulating, we directly count:

sum(x < k for x in nums)
= sum(3 < 5, 7 < 5, 2 < 5, 9 < 5, 1 < 5, 5 < 5)
= sum(True, False, True, False, True, False)
= sum(1, 0, 1, 0, 1, 0)
= 3

The answer is 3 operations, which matches our simulation. This confirms that counting elements less than k directly gives us the minimum number of operations needed.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the code iterates through each element in the array exactly once to check if x < k, performing a constant-time comparison for each element.

The space complexity is O(1). Although the generator expression (x < k for x in nums) is used, the sum() function consumes it immediately without storing all the boolean values in memory. The function only maintains a running sum counter, which requires constant extra space regardless of the input size.

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

Common Pitfalls

1. Misunderstanding the Problem Requirements

Pitfall: Assuming we need to modify elements to make them ≥ k, rather than removing them entirely.

Some might incorrectly try to:

# WRONG: Trying to count operations to modify elements
def minOperations(self, nums: List[int], k: int) -> int:
    operations = 0
    for x in nums:
        if x < k:
            operations += (k - x)  # Wrong! We're not incrementing values
    return operations

Solution: Remember that each operation removes one occurrence of the smallest element. We simply need to count how many elements are less than k, not calculate how much to increase them.

2. Edge Case: Empty Array

Pitfall: Not handling the case where nums is empty or has zero length.

# Potential issue if not careful with edge cases
def minOperations(self, nums: List[int], k: int) -> int:
    if not nums:  # This check might be forgotten
        return 0
    return sum(x < k for x in nums)

Solution: The provided solution actually handles this gracefully - sum() on an empty generator returns 0, which is correct.

3. Incorrect Comparison Operator

Pitfall: Using <= instead of < when counting elements.

# WRONG: Using less than or equal to
def minOperations(self, nums: List[int], k: int) -> int:
    return sum(x <= k for x in nums)  # Wrong! This includes k itself

Solution: Use strict inequality < since we want remaining elements to be greater than or equal to k, meaning we only remove elements strictly less than k.

4. Attempting to Sort Unnecessarily

Pitfall: Sorting the array first, thinking it's needed to identify smallest elements.

# INEFFICIENT: Unnecessary sorting
def minOperations(self, nums: List[int], k: int) -> int:
    nums.sort()  # O(n log n) - unnecessary!
    count = 0
    for x in nums:
        if x < k:
            count += 1
        else:
            break
    return count

Solution: Since we're counting all elements less than k regardless of order, sorting adds unnecessary O(n log n) complexity. The O(n) traversal is sufficient.

5. Type Assumptions

Pitfall: Not considering that array elements or k could be negative numbers.

# Might fail with negative numbers if special logic was assumed
def minOperations(self, nums: List[int], k: int) -> int:
    # If someone incorrectly assumes all values are positive
    if k <= 0:  # Unnecessary constraint
        return len(nums)
    return sum(x < k for x in nums)

Solution: The straightforward comparison x < k works correctly for all integers, including negative numbers. No special handling is needed.