<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:
        presum = [nums[0]]
        for i in range(1,len(nums)):
            presum.append(presum[-1]+nums[i])
        dp = [[-1 for _ in range(k)] for _ in range(len(nums))]
        
        def backtrack(partRemaining,index):
            #base condition
            if partRemaining<0:
                return 0
            if partRemaining==0:
                return presum[len(presum)-1]-presum[index]+nums[index]
            if index==len(nums):
                return 0

            if dp[index][partRemaining]!=-1:
                return dp[index][partRemaining]

            
            ans = [1e9]
            if partRemaining == k-1:
                for i in range(len(nums)-partRemaining):
                    ans.append(max(presum[i],backtrack(partRemaining-1,i+1)))
            else:
                for i in range(index,len(nums)-partRemaining):
                    #print(presum[i]-presum[index]+nums[index],index,i)
                    ans.append(max(presum[i]-presum[index]+nums[index],backtrack(partRemaining-1,i+1)))

            dp[index][partRemaining] = min(ans)
            return dp[index][partRemaining]
        
        return backtrack(k-1,0)

17600