<p>Given an integer array <code>nums</code> and an integer <code>k</code>, split <code>nums</code> into <code>k</code> non-empty subarrays such that the largest sum of any subarray is <strong>minimized</strong>.</p>

<p>Return <em>the minimized largest sum of the split</em>.</p>

<p>A <strong>subarray</strong> is a contiguous part of the array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,2,5,10,8], k = 2
<strong>Output:</strong> 18
<strong>Explanation:</strong> There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8], where the largest sum among the two subarrays is only 18.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5], k = 2
<strong>Output:</strong> 9
<strong>Explanation:</strong> There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= min(50, nums.length)</code></li>
</ul>

class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        n = len(nums)
        prefixSum = [0] * n
        prefixSum[0] = nums[0]
        for i in range(1, n):
            prefixSum[i] = prefixSum[i - 1] + nums[i]
        memo = [[sys.maxsize] * n for _ in range(k + 1)]

        def dp(remainK, end):
            if remainK == 1:
                return prefixSum[end]
            elif remainK > end + 1:
                return sys.maxsize
            elif memo[remainK][end] != sys.maxsize:
                return memo[remainK][end]
            else:
                ans = sys.maxsize
                for i in range(end):
                    ans = min(ans, max(dp(remainK - 1, i), prefixSum[end] - prefixSum[i]))
                memo[remainK][end] = ans
                return ans

        return dp(k, n - 1)
18500