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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1049" 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 any two 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 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 smallest possible weight of the left stone. If there are no stones left, return 0.   Example 1:  Input: stones = [2,7,4,1,8,1] Output: 1 Explanation: We can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then, we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then, we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then, we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.  Example 2:  Input: stones = [31,26,33,21,40] Output: 5    Constraints:  1 <= stones.length <= 30 1 <= stones[i] <= 100  

	# Explanation
	Here's the approach to solve this problem:

*   **Dynamic Programming:** The core idea is to use dynamic programming to determine the possible sums we can achieve by partitioning the stones into two groups. The goal is to find a partition such that the difference between the sums of the two groups is minimized.
*   **Subset Sum Problem:** The problem is essentially reduced to a subset sum problem. We try to find a subset of stones whose sum is as close as possible to half the total sum of all stones.
*   **Optimization:** By finding this optimal subset, we effectively minimize the weight of the remaining stone, which is the difference between the total sum and twice the subset sum.

*   **Runtime Complexity:** O(n\*sum), where n is the number of stones and sum is the total sum of the stones.  **Storage Complexity:** O(sum).

	
	# Code
	```python
	def lastStoneWeightII(stones):
    """
    Finds the smallest possible weight of the left stone after smashing all stones.

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

    Returns:
        The smallest possible weight of the left stone.
    """

    total_sum = sum(stones)
    n = len(stones)
    target = total_sum // 2

    # dp[i] is True if a subset of the stones can sum up to i
    dp = [False] * (target + 1)
    dp[0] = True

    for stone in stones:
        for i in range(target, stone - 1, -1):
            dp[i] = dp[i] or dp[i - stone]

    # Find the largest sum achievable that is less than or equal to target
    for i in range(target, -1, -1):
        if dp[i]:
            return total_sum - 2 * i

    return 0
	```