<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)
        prefix = [0 for _ in range(n+1)]
        for i in range(1,n+1):
            prefix[i] = prefix[i-1] + nums[i-1]
        memo = {}
        def dfs(start, splits):
            if start==n:
                if splits==k:
                    return 0
                return float("inf")
            if (start, splits) in memo:
                return memo[start, splits]
            if splits==k-1:
                return prefix[n]-prefix[start]
            res = float("inf")
            for i in range(start+1, n+1):
                if prefix[i] - prefix[start] > res:
                    break
                tmp = max(prefix[i]-prefix[start], dfs(i, splits+1))
                res = min(res, tmp)
            memo[start, splits] = res
            return res
        return dfs(0, 0)



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