Problem Description

You have an array of integers nums (0-indexed) and an integer k. Your goal is to make all elements in the array greater than or equal to k through a series of operations.

In each operation, you can:

1. Select the two smallest integers x and y from the array
2. Remove both x and y from the array
3. Calculate a new value using the formula: min(x, y) * 2 + max(x, y)
4. Insert this new value back into the array at any position

You can only perform an operation if the array contains at least 2 elements.

The task is to find the minimum number of operations needed to ensure every element in the array is greater than or equal to k.

For example, if you have nums = [2, 11, 10, 1, 3] and k = 10:

• The two smallest elements are 1 and 2
• Remove them and insert 1 * 2 + 2 = 4
• Continue this process until all remaining elements are ≥ 10

The solution uses a min heap to efficiently track and extract the smallest elements. It repeatedly takes the two smallest values, combines them using the given formula, and puts the result back into the heap. This continues until either there's only one element left or the smallest element is already ≥ k.

Intuition

The key insight is that we want to eliminate all elements smaller than k while minimizing the number of operations. Since each operation combines two elements into one, we're essentially reducing the array size by one each time.

When we combine two numbers using the formula min(x, y) * 2 + max(x, y), the resulting value is always larger than both original values. This means we're gradually increasing the values in our array.

The greedy strategy emerges from observing that we should always combine the two smallest elements. Why? Because:

1. We must eventually deal with all elements smaller than k
2. Combining the two smallest elements gives us the best chance of creating a value that's still manageable and won't unnecessarily inflate larger values
3. If we combined a small element with a large element, we'd get an even larger result, which doesn't help us minimize operations

Since we repeatedly need to find and extract the two smallest elements, a min heap is the perfect data structure. It gives us O(log n) access to the smallest element and maintains the sorted property as we add new combined values back.

The process naturally terminates when either:

• The smallest element in our heap is already ≥ k (mission accomplished)
• We only have one element left (can't perform more operations)

This greedy approach of always combining the two smallest elements guarantees the minimum number of operations because we're processing elements in the most efficient order - dealing with the problematic small values first while keeping the resulting values as small as possible.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The solution uses a min heap (priority queue) to efficiently manage the array elements and perform the required operations.

Step-by-step implementation:

1. Initialize the heap: Convert the input array nums into a min heap using heapify(nums). This operation arranges the array so that the smallest element is always at index 0, and takes O(n) time.

2. Set up the counter: Initialize ans = 0 to track the number of operations performed.

3. Main loop: Continue operations while two conditions are met:

   • The heap contains at least 2 elements (len(nums) > 1)
   • The smallest element is less than k (nums[0] < k)

4. Perform each operation:

   • Extract the two smallest elements using heappop(nums) twice. Let's call them x and y
   • Calculate the new value: Since x is popped first, it's guaranteed to be ≤ y, so the formula simplifies to x * 2 + y
   • Insert the new value back into the heap using heappush(nums, x * 2 + y)
   • Increment the operation counter: ans += 1

5. Return the result: Once the loop exits (either because all elements are ≥ k or only one element remains), return ans

Time Complexity: O(n log n) where n is the initial size of the array. In the worst case, we might need to perform operations on almost all elements, and each heap operation (pop and push) takes O(log n) time.

Space Complexity: O(1) as we're modifying the input array in-place (the heap is built on the original array).

The beauty of this approach is that the heap automatically maintains the order we need - we can always access the two smallest elements in O(log n) time, making the greedy strategy efficient to implement.

Example Walkthrough

Let's walk through a concrete example with nums = [1, 5, 3, 9] and k = 7.

Initial Setup:

• Convert array to min heap: [1, 5, 3, 9] (1 is at the root)
• Operations counter: ans = 0

Operation 1:

• Check conditions: heap has 4 elements (> 1) ✓, smallest element is 1 (< 7) ✓
• Extract two smallest: x = 1 (first pop), y = 3 (second pop)
• Heap after extraction: [5, 9]
• Calculate new value: 1 * 2 + 3 = 5
• Insert back into heap: [5, 9, 5] → heap rearranges to [5, 9, 5]
• Increment counter: ans = 1

Operation 2:

• Check conditions: heap has 3 elements (> 1) ✓, smallest element is 5 (< 7) ✓
• Extract two smallest: x = 5 (first pop), y = 5 (second pop)
• Heap after extraction: [9]
• Calculate new value: 5 * 2 + 5 = 15
• Insert back into heap: [9, 15] → heap maintains [9, 15]
• Increment counter: ans = 2

Loop Termination:

• Check conditions: heap has 2 elements (> 1) ✓, but smallest element is 9 (≥ 7) ✗
• Loop exits as all elements are now ≥ 7

Final Result: ans = 2

The array transformed from [1, 5, 3, 9] to [9, 15] in 2 operations, with all elements now greater than or equal to 7.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The initial heapify(nums) operation takes O(n) time to build a min-heap from the array. In the worst case, the while loop can run up to n-1 times (when we need to merge all elements into one). Each iteration performs:

• Two heappop() operations: O(log n) each
• One heappush() operation: O(log n)

Since each iteration has O(log n) complexity and we can have up to O(n) iterations, the overall time complexity is O(n × log n).

Space Complexity: O(n)

The space complexity is O(n) where n is the length of the array. The heapify() operation modifies the input list in-place to create a heap structure, requiring O(1) additional space. However, since we're considering the total space used including the input, the space complexity is O(n) for storing the heap structure of the input array.

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

Common Pitfalls

1. Not Checking if the Goal is Achievable

The most critical pitfall is assuming that it's always possible to make all elements ≥ k. If the array initially contains any element that equals 0, the goal becomes impossible to achieve.

Why this happens: When we have a 0 in the array, any operation involving it will produce:

• 0 * 2 + y = y (if 0 is the smaller element)
• x * 2 + 0 = 2x (if 0 is the larger element)

If we start with two zeros, we get 0 * 2 + 0 = 0, creating an infinite loop where we can never increase the minimum value.

Solution:

def minOperations(self, nums: List[int], k: int) -> int:
    # Check if any element is 0 and k > 0
    if 0 in nums and k > 0:
        return -1  # Impossible to achieve
  
    heapify(nums)
    operations_count = 0
  
    while len(nums) > 1 and nums[0] < k:
        first_min = heappop(nums)
        second_min = heappop(nums)
        combined_value = first_min * 2 + second_min
        heappush(nums, combined_value)
        operations_count += 1
  
    # Final check: if we're left with one element less than k
    if nums and nums[0] < k:
        return -1  # Cannot achieve the goal
  
    return operations_count

2. Modifying the Original Input Array

The solution modifies the input array directly by converting it into a heap. This can be problematic if the caller expects the original array to remain unchanged.

Solution:

def minOperations(self, nums: List[int], k: int) -> int:
    # Create a copy to avoid modifying the original array
    heap = nums.copy()
    heapify(heap)
  
    operations_count = 0
  
    while len(heap) > 1 and heap[0] < k:
        first_min = heappop(heap)
        second_min = heappop(heap)
        combined_value = first_min * 2 + second_min
        heappush(heap, combined_value)
        operations_count += 1
  
    return operations_count

3. Integer Overflow in Other Languages

While Python handles arbitrary-precision integers automatically, implementing this solution in languages like Java or C++ could lead to integer overflow when repeatedly combining values.

Why this matters: The formula min * 2 + max can grow quickly. After multiple operations, the combined values might exceed the maximum integer limit (e.g., 2^31 - 1 for 32-bit integers).

Solution for other languages:

• Use long or long long data types
• Add overflow checking
• Consider the maximum possible value after all operations

4. Assuming Elements are Already Sorted

Some might try to optimize by assuming the input is sorted or by sorting it first, then using indices instead of a heap. This approach fails because after each operation, the new combined value might not maintain the sorted order.

Wrong approach:

# This doesn't work correctly!
nums.sort()
operations = 0
while len(nums) > 1 and nums[0] < k:
    new_val = nums[0] * 2 + nums[1]
    nums = nums[2:] + [new_val]  # Inefficient and incorrect
    nums.sort()  # O(n log n) per operation!
    operations += 1

The heap-based solution is correct because it efficiently maintains the min-heap property after each insertion, ensuring we always get the true two smallest elements.