Solution Approach

The solution implements the greedy strategy using Python's built-in heapq module, which provides a min heap implementation.

Step 1: Initialize the Min Heap

heapify(sticks)

We transform the input array sticks into a min heap in-place. This operation takes O(n) time and ensures the smallest element is always at index 0.

Step 2: Initialize Cost Counter

ans = 0

We maintain a variable ans to track the cumulative cost of all connections.

Step 3: Connect Sticks Until One Remains

while len(sticks) > 1:

We continue the process as long as there are at least 2 sticks to connect.

Step 4: Extract and Connect Two Smallest Sticks

z = heappop(sticks) + heappop(sticks)

In each iteration:

• heappop(sticks) removes and returns the smallest stick (first call)
• heappop(sticks) removes and returns the second smallest stick (second call)
• We add their lengths to get the new stick length z

Step 5: Accumulate Cost

ans += z

The cost of this connection (which equals the length of the new stick) is added to our total cost.

Step 6: Insert New Stick Back

heappush(sticks, z)

The newly formed stick is inserted back into the heap, maintaining the heap property.

Time Complexity: O(n log n) where n is the number of sticks. We perform approximately n-1 iterations, and each iteration involves two heappop operations and one heappush operation, each taking O(log n) time.

Space Complexity: O(1) extra space since we modify the input array in-place (not counting the space used by the input itself).

Example Walkthrough

Let's walk through the solution with sticks = [1, 8, 3, 5]:

Initial State:

• Array: [1, 8, 3, 5]
• After heapify: [1, 3, 8, 5] (min heap structure)
• Total cost: 0

Iteration 1:

• Extract two smallest: 1 and 3
• Connect them: 1 + 3 = 4
• Add cost: total = 0 + 4 = 4
• Insert back: heap becomes [4, 5, 8]

Iteration 2:

• Extract two smallest: 4 and 5
• Connect them: 4 + 5 = 9
• Add cost: total = 4 + 9 = 13
• Insert back: heap becomes [8, 9]

Iteration 3:

• Extract two smallest: 8 and 9
• Connect them: 8 + 9 = 17
• Add cost: total = 13 + 17 = 30
• Insert back: heap becomes [17]

Final Result:

• Only one stick remains
• Minimum total cost = 30

The greedy approach ensures we always combine the smallest sticks first, minimizing how many times larger values contribute to the total cost. Notice how the stick of length 8 (initially the largest) only participated in one connection operation, while smaller sticks like 1 and 3 had their values included multiple times through the combined sticks they formed.

Time and Space Complexity

The time complexity is O(n × log n), where n is the length of the array sticks.

• The initial heapify(sticks) operation takes O(n) time to build a min-heap from the array.
• The while loop runs n - 1 times (until only one stick remains).
• In each iteration, we perform:
  • Two heappop() operations: O(log n) each
  • One heappush() operation: O(log n)
  • Total per iteration: O(log n)
• Since we have n - 1 iterations, the loop contributes O(n × log n) time.
• Overall time complexity: O(n) + O(n × log n) = O(n × log n)

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

• The heapify() operation modifies the input array in-place and uses O(1) extra space.
• The heap operations (heappop and heappush) work on the existing array without requiring additional space proportional to the input size.
• The space is dominated by the input array itself, which has size n.
• Therefore, the space complexity is O(n).

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

Common Pitfalls

1. Edge Case: Single or Empty Stick Array

Pitfall: The code doesn't explicitly handle the case where the input contains only one stick or is empty. While the current implementation works correctly (returning 0 for a single stick), it's not immediately obvious why.

Why it happens: When len(sticks) == 1, the while loop never executes, and total_cost remains 0. This is correct since no connections are needed, but it's implicit rather than explicit.

Solution:

def connectSticks(self, sticks: List[int]) -> int:  # Explicitly handle edge cases for clarity  if len(sticks) <= 1:  return 0   heapify(sticks)  total_cost = 0  # ... rest of the code

2. Using Wrong Heap Type (Max Heap Instead of Min Heap)

Pitfall: Attempting to use a max heap or forgetting that Python's heapq is a min heap by default. Some might try to negate values to simulate a max heap, which would give incorrect results for this problem.

Why it happens: In some problems, we need a max heap and use negation as a workaround. Applying this pattern here would cause us to connect the largest sticks first, resulting in suboptimal cost.

Incorrect approach example:

# WRONG: Don't negate values for this problem! heapify([-x for x in sticks]) # This would create wrong behavior

Solution: Use the min heap directly as shown in the original code. The algorithm requires connecting smallest sticks first.

3. Modifying Original Input Without Consideration

Pitfall: The solution modifies the input array sticks in-place. If the caller expects the original array to remain unchanged, this causes unexpected side effects.

Why it happens: Using heapify(sticks) modifies the original list structure. After the function returns, the original sticks array is altered.

Solution: Create a copy if preserving the original is important:

def connectSticks(self, sticks: List[int]) -> int:  # Create a copy to avoid modifying the original  sticks_heap = sticks.copy() # or sticks[:]  heapify(sticks_heap)   total_cost = 0  while len(sticks_heap) > 1:  combined_length = heappop(sticks_heap) + heappop(sticks_heap)  total_cost += combined_length  heappush(sticks_heap, combined_length)   return total_cost

4. Inefficient Two-Line Pop Operations

Pitfall: Some might write the extraction of two sticks in separate lines with intermediate variables when not needed for clarity.

Less efficient style:

first = heappop(sticks) second = heappop(sticks) combined = first + second

Why it matters: While functionally correct, it uses more lines and variables than necessary. The original compact form z = heappop(sticks) + heappop(sticks) is more concise.

Note: This is more about style than correctness. Choose based on team preferences for readability vs. conciseness.