<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:
        l = []
        s = 0
        for i in nums:
            s+=i
            l.append(s)

        @cache
        def f(idx, num_splits):
            if num_splits > idx + 1:
                return float('inf')

            if idx==0 or num_splits==1:
                return l[idx]
            result = float('inf')
            i = num_splits - 1
            j = idx

            while i<j:
                cut = (i+j)//2
                last_sum = l[idx] - l[cut-1]
                prev_split = f(cut - 1, num_splits - 1)
                result=min(result, max(prev_split, last_sum))
                if last_sum>prev_split:
                    i=min(cut+1, j)
                elif last_sum==prev_split:
                    i = cut
                    break
                else:
                    j=max(cut-1, i)

            cut = i
            last_sum = l[idx] - l[cut-1]
            prev_split = f(cut - 1, num_splits - 1)
            result=min(result, max(prev_split, last_sum))
            return result

        return f(len(nums)-1, k)


            
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