Problem Description

You are given an array sticks where each element represents the length of a stick (positive integer).

You can connect any two sticks together to form a single stick. When you connect two sticks with lengths x and y, the resulting stick has length x + y, and you must pay a cost equal to x + y.

Your goal is to connect all the sticks together until only one stick remains. You need to find the minimum total cost to achieve this.

For example, if you have sticks with lengths [2, 4, 3]:

• You could first connect sticks of length 2 and 3, paying cost 5, resulting in sticks [5, 4]
• Then connect the remaining two sticks of length 5 and 4, paying cost 9
• Total cost would be 5 + 9 = 14

The strategy uses a greedy approach with a min heap. By always connecting the two shortest sticks first, we minimize the cost at each step. The algorithm repeatedly:

1. Extracts the two smallest sticks from the heap
2. Connects them (calculating the cost)
3. Adds the cost to the running total
4. Inserts the new combined stick back into the heap
5. Continues until only one stick remains

Intuition

The key insight is that when we connect two sticks, the resulting stick will potentially be used in future connections. Since we pay the sum of lengths each time we connect, longer sticks that participate in multiple connections will contribute their length to the total cost multiple times.

Think of it this way: if we have a very long stick, we want to use it as few times as possible in our connections. Conversely, shorter sticks should be combined early because their smaller values will be added to the total cost multiple times as they form parts of larger sticks.

Consider why connecting the two shortest sticks is optimal at each step:

• When we connect sticks of length a and b, we pay a + b
• The resulting stick of length a + b might be used again in future connections
• If we had instead connected a short stick with a long stick early, that long stick's value would be included in more subsequent operations, increasing the total cost

This naturally leads to a greedy strategy: always connect the two shortest available sticks. By doing this, we ensure that:

1. Smaller values are "promoted" to larger sticks early
2. Larger values participate in fewer connection operations
3. The overall cost is minimized

A min heap (priority queue) is the perfect data structure for this approach because it allows us to efficiently:

• Extract the two minimum elements in O(log n) time
• Insert the newly formed stick back in O(log n) time
• Always maintain access to the current shortest sticks

Learn more about Greedy and Heap (Priority Queue) patterns.

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.

Solution Implementation

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.