<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:
        # monotic: the largest sum is smaller, the splits needed will be bigger
        r = sum(nums)
        l = max(nums)

        # we need to decide if it can split within k subarrays for each 
        # possible max sum

        def canSplit(maxSum):
            current_sum = 0
            subarrays = 1
            
            for num in nums:
                current_sum += num
                if current_sum > maxSum:
                    subarrays += 1
                    current_sum = num  # Start a new subarray with the current
                    if subarrays > k:
                        return False
            return True


        while l < r:
            mid = (l+r)//2
            if canSplit(mid): # satisfy requirement, try smaller largest sum
                r = mid
            else:
                l = mid + 1
        return l




        
                





        
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