# Solving Leetcode Interviews in Seconds with AI: Last Stone Weight


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1046" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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, and If x != y, the stone of weight x is destroyed, and the stone of weight y has 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:  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.  Example 2:  Input: stones = [1] Output: 1    Constraints:  1 <= stones.length <= 30 1 <= stones[i] <= 1000  

	# Explanation
	Here's an efficient solution to the stone smashing problem:

*   **Key Idea:** Use a max-heap (priority queue) to efficiently find the two heaviest stones at each step.
*   **Simulation:** Simulate the stone smashing process by repeatedly extracting the two largest stones, calculating the difference (if any), and re-inserting the result back into the heap.
*   **Heap Data Structure:** Python's `heapq` module provides a min-heap implementation. We'll negate the stone weights to effectively simulate a max-heap.

*   **Runtime Complexity:** O(N + M log N), where N is the initial number of stones and M is the total number of operations performed (smashing). Since M is bounded by N, the overall runtime is O(N log N).
    **Storage Complexity:** O(N) to store the stones in the heap.

	
	# Code
	```python
	import heapq

def lastStoneWeight(stones):
    """
    Finds the weight of the last remaining stone after smashing stones.

    Args:
        stones: A list of integers representing the weights of the stones.

    Returns:
        The weight of the last remaining stone, or 0 if no stones are left.
    """
    stones = [-stone for stone in stones]  # Negate for max-heap simulation
    heapq.heapify(stones)

    while len(stones) > 1:
        stone1 = -heapq.heappop(stones)
        stone2 = -heapq.heappop(stones)

        if stone1 != stone2:
            heapq.heappush(stones, -(stone1 - stone2))

    return -stones[0] if stones else 0
	```