<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> consisting of <code>3 * n</code> elements.</p>

<p>You are allowed to remove any <strong>subsequence</strong> of elements of size <strong>exactly</strong> <code>n</code> from <code>nums</code>. The remaining <code>2 * n</code> elements will be divided into two <strong>equal</strong> parts:</p>

<ul>
	<li>The first <code>n</code> elements belonging to the first part and their sum is <code>sum<sub>first</sub></code>.</li>
	<li>The next <code>n</code> elements belonging to the second part and their sum is <code>sum<sub>second</sub></code>.</li>
</ul>

<p>The <strong>difference in sums</strong> of the two parts is denoted as <code>sum<sub>first</sub> - sum<sub>second</sub></code>.</p>

<ul>
	<li>For example, if <code>sum<sub>first</sub> = 3</code> and <code>sum<sub>second</sub> = 2</code>, their difference is <code>1</code>.</li>
	<li>Similarly, if <code>sum<sub>first</sub> = 2</code> and <code>sum<sub>second</sub> = 3</code>, their difference is <code>-1</code>.</li>
</ul>

<p>Return <em>the <strong>minimum difference</strong> possible between the sums of the two parts after the removal of </em><code>n</code><em> elements</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,2]
<strong>Output:</strong> -1
<strong>Explanation:</strong> Here, nums has 3 elements, so n = 1. 
Thus we have to remove 1 element from nums and divide the array into two equal parts.
- If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
- If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
- If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.
The minimum difference between sums of the two parts is min(-1,1,2) = -1. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,9,5,8,1,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.
If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.
To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.
It can be shown that it is not possible to obtain a difference smaller than 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>nums.length == 3 * n</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class KLargest:
    def __init__(self, kElementList):
        self.k = len(kElementList)
        self.heap = kElementList
        heapq.heapify(self.heap)
        self.curSum = sum(self.heap)

    def add(self, x):
        if x > self.heap[0]:
            self.curSum -= self.heap[0]
            self.curSum += x
            heapq.heapreplace(self.heap, x)
        
    def sum(self):
        return self.curSum

class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        n_ = len(nums) // 3

        # We want to maximize the second sum while minimizing the first sum
        # Can we be greedy?
        # Consider using min/max heaps for the left/right
        # then just greedily removing the element that minimizes the distance the most

        # left -> small
        # right -> BIG

        # Consider: There is some point (within the middle third) that we can split the array. 
        # Any elements to the left are left sum, and any elemnts to the right are right sum
        # The amount of removals on each side is necessarily determined by the location of the partition
        # and you would simply greedily remove the smallest/largest elements there

        # You can imagine sliding a window over from the left
        # and adding elements to the left
        # And simply maintain the n smallest elements
        # This same process can also be done from the right

        # Minimium sum on left at each possible splitting point.
        minSumFromLeft = [0] * (n_ + 1)
        leftSmallest = KLargest([-x for x in nums[:n_]])
        minSumFromLeft[0] = -leftSmallest.sum()
        for i in range(n_):
            leftSmallest.add(-nums[n_ + i])
            minSumFromLeft[i+1] = -leftSmallest.sum()
        del leftSmallest

        # This need not be stored in an array, Im just doing that because it makes it easier to reason with right now.
        # Maximum sum on right at each possible splitting point
        rightLargest = KLargest(nums[-n_:])
        out = minSumFromLeft[n_] - rightLargest.sum()
        for i in range(n_-1, -1, -1):
            rightLargest.add(nums[n_ + i])
            out = min(out, minSumFromLeft[i] - rightLargest.sum())

        return out

        
40700