Minimum Cost to Connect Sticks

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Greedy AlgorithmsArraysDynamic Programming

You have some sticks with positive integer lengths. These lengths are given as an array sticks, where sticks[i] is the length of the i^th stick.

You can connect any two sticks of lengths x and y into one stick by paying a cost of x + y. You perform this action until there is one stick remaining.

Return the minimum cost of connecting all the given sticks into one stick in this way.

Example 1:

Input: sticks = [2,4,3]
Output: 14
Explanation: You start with sticks = [2,4,3].
1. Combine sticks 2 and 3 for a cost of 2 + 3 = 5. Now you have sticks = [4,5].
2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have one stick = [9].
The total cost is 5 + 9 = 14.

Example 2:

Input: sticks = [1,8,3,5]
Output: 30
Explanation: You start with sticks = [1,8,3,5].
1. Combine sticks 1 and 3 for a cost of 1 + 3 = 4. Now you have sticks = [4,5,8].
2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have sticks = [8,9].
3. Combine sticks 8 and 9 for a cost of 8 + 9 = 17. Now you have one stick = [17].
The total cost is 4 + 9 + 17 = 30.

Example 3:

Input: sticks = [5]
Output: 0
Explanation: There is only one stick, so you don't need to do anything and the total cost is 0.

Constraints:

1 <= sticks.length <= 10⁴
1 <= sticks[i] <= 10⁴

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the length of each stick, and the number of sticks?
Can the input array of stick lengths be empty, or contain zero or negative values?
Is the goal to minimize the *total* cost, or is there some other cost function?
If there's only one stick, what should the returned cost be?
Are we guaranteed that the stick lengths are integers, or might they be floating-point numbers?

Brute Force Solution

Approach

The brute force approach to connecting sticks involves exploring every possible combination of stick pairings to find the lowest cost. We will try out all possible groupings, calculating the cost for each arrangement. Finally, we compare all the total costs and find the absolute minimum.

Here's how the algorithm would work step-by-step:

Consider all possible pairs of sticks that can be combined first.
For each of these pairs, calculate the cost of combining them (which is the sum of their lengths).
After combining a pair, we have one less stick.
Repeat the process by considering all possible pairs of the new set of sticks (including the combined one).
Keep calculating the cost of each combination and remember the total cost of all combinations made so far for each possible pairing path.
Continue until there is only one stick remaining; at this point we know the total cost for that particular path of combinations.
Do this for every single possible path you can take to combine the sticks.
Finally, compare the total costs of all paths, and select the path that had the minimum total cost.

Code Implementation

def minimum_cost_to_connect_sticks_brute_force(sticks):
    minimum_total_cost = float('inf')

    def calculate_cost(current_sticks, current_total_cost):
        nonlocal minimum_total_cost

        # Base case: only one stick left
        if len(current_sticks) == 1:
            minimum_total_cost = min(minimum_total_cost, current_total_cost)
            return

        # Iterate through all possible pairs of sticks
        for first_stick_index in range(len(current_sticks)):
            for second_stick_index in range(first_stick_index + 1, len(current_sticks)):

                # Form the new set of sticks after combining two sticks
                new_sticks = []
                for index in range(len(current_sticks)):
                    if index != first_stick_index and index != second_stick_index:
                        new_sticks.append(current_sticks[index])

                # The cost of combining the pair of sticks.
                combined_stick_length = current_sticks[first_stick_index] + current_sticks[second_stick_index]
                new_sticks.append(combined_stick_length)

                # Recursively calculate the cost of the remaining sticks
                calculate_cost(new_sticks, current_total_cost + combined_stick_length)

    # Start the recursive calculation with the initial sticks and a cost of 0.
    calculate_cost(sticks, 0)

    return minimum_total_cost

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach explores every possible combination of stick pairings. With n sticks, there are n-1 possible initial pairings. After each pairing, the number of sticks reduces, leading to a factorial-like behavior in the number of possible combination paths. Therefore, the algorithm essentially enumerates all possible ways to combine the sticks, resembling the permutations, leading to O(n!) time complexity as we explore every possible path in combining the sticks.

Space Complexity

O(N!) – The described brute force approach explores all possible pairings of sticks. The number of possible pairing paths grows factorially with the number of sticks, N. To explore all paths, the algorithm must keep track of the intermediate stick configurations and costs associated with each path; this implies storing a tree-like structure, where each branch represents a sequence of stick combinations. In the worst case, this leads to storing information for every possible combination path, resulting in space usage proportional to the number of possible combinations, which is O(N!).

Optimal Solution

Approach

The core idea is to always combine the smallest sticks first. This avoids having small sticks contribute to the cost multiple times during the overall connection process. By repeatedly merging the two shortest sticks, we ensure the accumulated cost is minimized at each stage.

Here's how the algorithm would work step-by-step:

Imagine you have a pile of sticks with different lengths.
Find the two shortest sticks in the pile.
Connect these two sticks together. The cost of this connection is the sum of their lengths.
Now, you have a new stick that's the combined length of the two you just connected. Put this new, longer stick back into the pile.
Repeat this process of finding the two shortest sticks, connecting them, and adding the new stick back to the pile until you are left with only one stick.
Keep track of the cost of each connection you make along the way. The total cost will be the sum of all the connection costs.

Code Implementation

import heapq

def minimum_cost_to_connect_sticks(stick_lengths):
    heapq.heapify(stick_lengths)
    total_cost = 0

    while len(stick_lengths) > 1:
        # Extract two shortest sticks.
        shortest_stick_one = heapq.heappop(stick_lengths)
        shortest_stick_two = heapq.heappop(stick_lengths)

        combined_stick_length = shortest_stick_one + shortest_stick_two
        total_cost += combined_stick_length

        # Add the combined stick back to the heap.
        heapq.heappush(stick_lengths, combined_stick_length)

    return total_cost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is repeatedly finding the two smallest sticks and merging them. To efficiently find the two smallest sticks at each step, we use a min-heap (priority queue). Inserting all n sticks into the min-heap takes O(n log n) time. Then, we perform (n-1) merge operations. Each merge involves extracting the two smallest elements (O(log n) each), summing them, and inserting the result back into the min-heap (O(log n)). Therefore, the (n-1) merge operations take a total of (n-1) * 3 * O(log n) = O(n log n) time. Since both the initial heap construction and the merges are O(n log n), the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm maintains a 'pile' of sticks which, in an efficient implementation, could be a min-heap. A min-heap built from N sticks will require storage proportional to the number of sticks. Thus, the auxiliary space required to store the min-heap is dependent on the number of input sticks, N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there are no sticks to connect and hence no cost.
Array with only one stick	Return 0, as a single stick requires no connections.
Array with two sticks	Return the sum of the two sticks, which is the only possible connection cost.
Large input array (performance considerations)	Use a min-heap data structure to efficiently find and combine the two smallest sticks in each step.
Input array contains very large integer values (potential overflow)	Use a data type (e.g., long in Java/C++) that can accommodate large sums without overflowing.
Input array contains zero values	Zero values should be treated as regular stick lengths and included in the cost calculation.
Input array contains negative values	Return an error or throw an exception, as stick lengths cannot be negative.
Array with many identical values	The min-heap handles duplicates correctly, selecting the smallest available values at each step to compute total cost.

Empty input array

How to Handle:

Return 0, as there are no sticks to connect and hence no cost.

Array with only one stick

How to Handle:

Return 0, as a single stick requires no connections.

Array with two sticks

How to Handle:

Return the sum of the two sticks, which is the only possible connection cost.

Large input array (performance considerations)

How to Handle:

Use a min-heap data structure to efficiently find and combine the two smallest sticks in each step.

Input array contains very large integer values (potential overflow)

How to Handle:

Use a data type (e.g., long in Java/C++) that can accommodate large sums without overflowing.

Input array contains zero values

How to Handle:

Zero values should be treated as regular stick lengths and included in the cost calculation.

Input array contains negative values

How to Handle:

Return an error or throw an exception, as stick lengths cannot be negative.

Array with many identical values

How to Handle:

The min-heap handles duplicates correctly, selecting the smallest available values at each step to compute total cost.