<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 Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        N = len(nums)
        n = N // 3
        first = []
        for i in range(n):
            heapq.heappush(first, [-nums[i], i])
        last = []
        for i in range(len(nums) - n, len(nums)):
            heapq.heappush(last, [nums[i], i])

        res = sum(nums)
        res = min(res, sum(nums[:n]) - sum(nums[-n:]))
        first_sum = sum(nums[:n])
        last_sum = sum(nums[-n:])
        changes = {}
        for i in range(n, len(nums) - n):
            cur_val = nums[i]
            top_val, top_index = first[0]
            top_val = -top_val
            if top_val > cur_val:
                changes[i] = top_index
                heapq.heappop(first)
                first_sum -= top_val 
                first_sum += cur_val
                heapq.heappush(first, [-cur_val, i])

        res = min(res, first_sum - last_sum)
        for i in range(len(nums) - n - 1, n - 1, -1):
            cur_val = nums[i]
            if i in changes:
                removed = nums[changes[i]]
                current = nums[i]
                first_sum -= current
                first_sum += removed
            
            top_val, top_index = last[0]
            if cur_val > top_val:
                last_sum += -top_val + cur_val
                heapq.heappop(last)
                heapq.heappush(last, [cur_val, i])
            
            res = min(res, first_sum - last_sum)

        return res



1676