<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>

# log(N)^k
# [....i....]

import copy 

class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        n = len(nums)

        sums = []
        c = 0
        for num in nums:
            c += num
            sums.append(c)

        mem = {}

        def dfs(i, kk):
            if kk == 1:
                return sums[-1] - sums[i] + nums[i]
            l, r = i, n - kk
            if l == r:
                return max(nums[i:])
            if (i, kk) in mem:
                return mem[(i, kk)]
            while l < r - 1:
                m = (l + r) // 2
                l_v = sums[m] - sums[i] + nums[i]
                r_v = dfs(m + 1, kk - 1)
                if l_v < r_v:
                    l = m
                else:
                    r = m
            l_v = sums[l] - sums[i] + nums[i]
            r_v = dfs(l + 1, kk - 1)

            res1 = max(sums[l] - sums[i] + nums[i], dfs(l + 1, kk - 1))
            res2 = max(sums[l + 1] - sums[i] + nums[i], dfs(l + 2, kk - 1))
            res = min(res1, res2)
            mem[(i, kk)] = res
            return min(res1, res2)

        res = dfs(0, k)
        return res
        
19300