Last Stone Weight Leetcode Problem 1046 [Python Solution]

In this blog post, we're going to tackle the Last Stone Weight problem from LeetCode.

This is an "Easy" level problem in the category of Heap/Priority Queue.

The problem involves an array of stones, and we are tasked with smashing the heaviest stones together until there is at most one stone left.

We'll provide a Python solution to this problem and explain the reasoning behind it.

Problem Overview

Question: You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones.

On each turn, we choose the heaviest two stones and smash them together.

Suppose the heaviest two stones have weights x and y with x <= y.

The result of this smash is:

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

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone.

If there are no stones left, return 0.

Example 1:

Input: stones = [2,7,4,1,8,1]

Output: 1

Explanation:
1. Combine 7 and 8 to get 1, so the array converts to [2,4,1,1,1].
2. Combine 2 and 4 to get 2, so the array converts to [2,1,1,1].
3. Combine 2 and 1 to get 1, so the array converts to [1,1,1].
4. Finally, combine 1 and 1 to get 0, so the array converts to [1], which is the value of the last stone.

Example 2:

Input: stones = [1]

Output: 1

Understanding Constraints

Before we dive into the solution, let's understand the constraints:

• 1 <= stones.length <= 30
• 1 <= stones[i] <= 1000

These constraints tell us that the input size is reasonably small, which means that an efficient solution should run in a reasonable time.

It's also important to note that there's no requirement to maintain the order of stones in the array; we are only concerned with their weights.

Last Stone Weight LeetCode Problem Solution

Let's explore different approaches to solving this problem.

1. Bruteforce Approach

One straightforward approach is to simulate the game as described in the problem statement.

We can repeatedly find the two heaviest stones, apply the smashing rules, and update the array until there's only one stone left.

Here's the Python code for the bruteforce approach:

def lastStoneWeight(stones):
    while len(stones) > 1:
        stones.sort()  # Sort the stones in ascending order.

x, y = stones.pop(), stones.pop()  # Get the two heaviest stones.

if x != y:
            stones.append(y - x)  # Add the remaining stone back.

return stones[0] if stones else 0

This approach works, but it's not very efficient.

The sort operation has a time complexity of O(n * log(n)), and we're doing this repeatedly.

So the overall time complexity of this solution is O(n^2 * log(n)).

2. An Efficient Approach using Max Heap

A more efficient approach is to use a max heap to keep track of the heaviest stones.

In Python, the heapq library provides a min heap, so we'll use negative values of the stone weights to simulate a max heap.

Here's the Python code for this efficient approach:

import heapq

def lastStoneWeight(stones):
    # Create a max heap by negating stone values.

stones = [-stone for stone in stones]
    heapq.heapify(stones)

    while len(stones) > 1:
        first = heapq.heappop(stones)
        second = heapq.heappop(stones)
        if second > first:
            heapq.heappush(stones, first - second)

    # If there's one stone left, return its positive value; otherwise, return 0. stones.append(0)
    return abs(stones[0])

In this efficient approach, we start by creating a max heap (negative values of stones) using the heapq library.

Then, in each iteration, we pop the two heaviest stones, calculate the difference, and push the result back into the heap.

We continue this process until there is only one stone left.

The overall time complexity of this solution is O(n * log(n)), which is significantly more efficient than the bruteforce approach.

3. Alternate Efficient Approach

There's also an alternate efficient approach using the _heapify_max method from the heapq library, which is not as commonly used but provides an interesting solution.

Here's the Python code for this alternate efficient approach:

import heapq

class Solution:
    def lastStoneWeight(self, stones):
        # Use _heapify_max to create a max heap.

heapq._heapify_max(stones)
        while len(stones) > 1:
            max_stone = heapq._heappop_max(stones)
            diff = max_stone - stones[0]
            if diff:
                heapq._heapreplace_max(stones, diff)
            else:
                heapq._heappop_max(stones)

        # If there's one stone left, return its value; otherwise, return 0. stones.append(0)
        return stones[0]

This approach uses the _heapify_max method to create a max heap, and the remaining logic is similar to the previous efficient approach.

The time complexity is still O(n * log(n)).

Time and Space Complexity

Let's summarize the time and space complexity of our efficient max heap approach:

• Time Complexity: O(n * log(n))
• Space Complexity: O(n)

The time complexity is dominated by the heap operations.

The space complexity is O(n) because we create a max heap to store the stone weights.

Reasoning Behind Our Approach

The reasoning behind our approach is quite simple.

We are given specific rules for smashing stones and need to find a way to efficiently implement these rules.

Using a max heap (or a min heap with negative values) allows us to always pick the heaviest stones, making the implementation straightforward and efficient.

We continue this process until there is at most one stone left.

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Conclusion

In this blog post, we tackled the Last Stone Weight problem from LeetCode.

We provided an efficient Python solution using a max heap (or min heap with negative values) to simulate the game and find the weight of the last remaining stone.

The time complexity of our approach is O(n * log(n)), making it an optimal solution for the problem.

We hope this guide has been helpful in understanding the problem and its solution.

If you have any questions, suggestions, or comments, please feel free to share them.

You can also check out the problem on LeetCode [here](https://

leetcode.com/problems/last-stone-weight/description/).

Remember that learning and practicing such coding problems is essential for improving your programming skills.

So, keep coding and happy problem-solving!