<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:
        total_sum=0
        dp = [ [-1]*len(nums) for _ in range(k+1) ]
        #dp = [[-1]*len(nums)]*(k+1)
        for i in range(len(nums)-1, -1, -1):
            total_sum+=nums[i]
            dp[1][i]=total_sum
        
        prefix_sum=[0]*(len(nums)+1)
        total_sum=0
        for i in range(len(nums)):
            total_sum+=nums[i]
            prefix_sum[i+1]=total_sum
        print(prefix_sum)
        
        for j in range(2, k+1):
            for i in range(len(nums)):
                for l in range(i+1, len(nums)):
                    currMax = max(prefix_sum[l]-prefix_sum[i], dp[j-1][l])
                    if(dp[j][i]==-1):
                        dp[j][i] = currMax
                    else:
                        dp[j][i] = min(dp[j][i], currMax)
        print(dp)
        return dp[k][0]
17800