human
I found an interesting problem on https://leetcode.com/problems/last-stone-weight-ii/:
We have a collection of rocks, each rock has a positive integer weight.
Each turn, we choose any two rocks 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 new weight y-x.

At the end, there is at most 1 stone left.  Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.)
 
Example 1:
Input: [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.

 
Note:

1 <= stones.length <= 30
1 <= stones[i] <= 100

I tried it in Python, but could not do it. Can you solve it?

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Write me a the Python program you requested?
assistant
#Here is the Python program you requested.

from heapq import heapify,heappush,heappop
class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        arr = stones
        n = len(stones)
        su = sum(arr)
        dp = [[0 for i in range(su+1)] for j in range(n+1)]
        for i in range(n + 1): 
            dp[i][0] = 1

        for j in range(1,n+1):
            for i in range(1,su+1):
                dp[j][i] = dp[j-1][i]
                if i-arr[j-1]>=0:
                    dp[j][i] |= dp[j-1][i-arr[j-1]]

        diff = sys.maxsize 

        # Find the largest j such that dp[n][j]  
        # is true where j loops from sum/2 t0 0  
        for j in range(su // 2, -1, -1): 
            if dp[n][j] == True: 
                diff = su - (2 * j) 
                break

        return diff