<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 dp(self, arr, start, k, subt, K, cache):
        if start >= len(arr):
            return float('inf')
        if k + 1 == K:
            return arr[-1] - subt
        if (start, k) in cache:
            return cache[(start, k)]
        best = float('inf')
        option = float('inf'), float('inf')
        for i in range(start, len(arr)):
            if (arr[i] - subt) == option[0]:
                # Window added a 0, which changes nothing
                continue
            option = arr[i] - subt, self.dp(arr, i + 1, k + 1, arr[i], K, cache)
            best = min(best, max(option))
        cache[(start, k)] = best
        return best

    def splitArray(self, nums: List[int], k: int) -> int:
        arr = [nums[0]]
        for n in nums[1:]:
            arr.append(arr[-1] + n)
        return self.dp(arr, 0, 0, 0, k, {})
        
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