Last Stone Weight in Python

We have a collection of stones, where each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

If x == y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has the new weight y-x.

At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

Example:

Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.

Understanding the Problem

The core challenge of this problem is to repeatedly find and process the two heaviest stones until only one stone remains or all stones are destroyed. This problem is significant in scenarios where we need to repeatedly process the largest elements, such as in certain scheduling or resource allocation problems.

Potential pitfalls include not efficiently finding the two heaviest stones, which can lead to suboptimal performance.

Approach

To solve this problem, we need to efficiently find and process the two heaviest stones. A naive approach would involve sorting the list repeatedly, but this is not optimal. Instead, we can use a max-heap (priority queue) to always access the heaviest stones efficiently.

Naive Solution

The naive solution involves sorting the list to find the two heaviest stones, which is inefficient due to the repeated sorting operations.

Optimized Solution

Using a max-heap allows us to efficiently find and remove the heaviest stones. In Python, we can use the `heapq` module, which provides a min-heap by default. By storing negative weights, we can simulate a max-heap.

Algorithm

Convert the list of stone weights into a max-heap by storing negative weights.
While there is more than one stone in the heap:
1. Extract the two heaviest stones (largest negative values).
2. If they are not equal, push the difference back into the heap.
If the heap is empty, return 0. Otherwise, return the weight of the remaining stone.

Code Implementation

import heapq

def lastStoneWeight(stones):
    # Convert all stone weights to negative to use heapq as a max-heap
    stones = [-stone for stone in stones]
    heapq.heapify(stones)
    
    # Process the stones until one or none is left
    while len(stones) > 1:
        # Extract the two heaviest stones
        first = heapq.heappop(stones)
        second = heapq.heappop(stones)
        
        # If they are not equal, push the difference back into the heap
        if first != second:
            heapq.heappush(stones, first - second)
    
    # If no stones are left, return 0; otherwise, return the weight of the remaining stone
    return -stones[0] if stones else 0

# Example usage
print(lastStoneWeight([2, 7, 4, 1, 8, 1]))  # Output: 1

Complexity Analysis

The time complexity of this approach is O(n log n), where n is the number of stones. This is because each insertion and extraction operation on the heap takes O(log n) time, and we perform these operations n-1 times. The space complexity is O(n) due to the storage of the heap.

Edge Cases

Potential edge cases include:

All stones having the same weight.
Only one stone in the input list.
No stones in the input list.

These cases are handled by the algorithm as it processes the heap until one or no stones are left.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases with a few stones.
Cases with all stones having the same weight.
Cases with only one stone.
Cases with no stones.

Using a testing framework like `unittest` in Python can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Think about data structures that can optimize the operations you need to perform.
Practice similar problems to improve your problem-solving skills.

Conclusion

Understanding and solving the "Last Stone Weight" problem helps improve your skills in using heaps and priority queues. It also highlights the importance of choosing the right data structures for efficient problem-solving.

Keep practicing and exploring further to enhance your algorithmic thinking and coding skills.

Last Stone Weight in Python (Time Complexity: O(n log n)) /homework