<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_sum = []
        curr_sum = 0
        if k == 1:
            return sum(nums)
        n = len(nums)
        for num in nums:
            curr_sum += num
            prefix_sum.append(curr_sum)
        @cache
        def dfs(i, k):
            if i >= n:
                return float('inf')
            
            if k == 1:
                return prefix_sum[n - 1] - (0 if i == 0 else prefix_sum[i - 1])
        
            minimum_largest_split_sum = prefix_sum[n- 1]
            for j in range(i, n - k + 1):
                curr_sum = prefix_sum[j] - (0 if i == 0 else prefix_sum[i - 1])
                max_sum = max(curr_sum, dfs(j + 1, k - 1))
                minimum_largest_split_sum = min(minimum_largest_split_sum, max_sum)
                if curr_sum >= minimum_largest_split_sum:
                    break
            
            
            return minimum_largest_split_sum
                
        return dfs(0, k)

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