<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)
        total = sum(nums)
        mem = {}
        prefix = [0] * (n+1)
        cur_sum = 0
        for i in range(1, n+1):
            cur_sum += nums[i-1]
            prefix[i] = cur_sum
        # print(f'{prefix}')
        def dfs(i: int, kk: int) -> int:
            if kk == 1:
                return prefix[n] - prefix[i]
            if (i, kk) in mem:
                return mem[(i, kk)]
            result = prefix[n]
            for j in range(i, n - kk + 1):
                first_sum = prefix[j+1] - prefix[i]
                largest_sum = max(first_sum, dfs(j+1, kk-1))
                result = min(result, largest_sum)
                if first_sum >= result:
                    break
            mem[(i, kk)] = result 
            return result
        
        return dfs(0, k)

        
23000