<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:
        prefix = [x for x in nums]
        for i in range(1, len(prefix)):
            prefix[i] = prefix[i-1] + prefix[i]

        def prefixSum(start, end):
            if start == 0: return prefix[end]
            return prefix[end] - prefix[start - 1]
            
        n = len(nums) 
        dp = {}
        for x in range(1, k+1):
            y = (n - 1) - (x - 1) + 1
            for i in range(0, y):
                if x == k and i > 0: break
                if x == 1: 
                    dp[(i, x)] =  prefixSum(i, n-1)
                else: 
                    curDp = inf 
                    for j in range(i, y):
                        curPrefixSum = prefixSum(i, j)
                        if curPrefixSum > curDp: continue
                        curMax = max(curPrefixSum, dp[(j+1, x-1)])
                        curDp = min(curDp, curMax)
                    dp[(i, x)] = curDp

        return dp[(0, k)]
25500