<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>

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

        min_idx = [0] * len(nums)

        for i in range(0, 2*n):
            if i < n-1:
                heapq.heappush(max_heap, -nums[i])
            elif i == n-1:
                heapq.heappush(max_heap, -nums[i])
                min_idx[i] = -sum(max_heap)
            else:
                min_idx[i] = min_idx[i-1] + nums[i]
                heapq.heappush(max_heap, -nums[i])
                min_idx[i] += heapq.heappop(max_heap)
        
        max_idx = [0] * len(nums)
        min_heap = []
        for i in range(len(nums)-2, n-2, -1):
            if i > 2*n-1:
                heapq.heappush(min_heap, nums[i+1])
            elif i == 2*n-1:
                heapq.heappush(min_heap, nums[i+1])
                max_idx[i] = sum(min_heap)
            else:
                max_idx[i] += max_idx[i+1] + nums[i+1]
                heapq.heappush(min_heap, nums[i+1])
                max_idx[i] -= heapq.heappop(min_heap)
        
        print(max_idx)
        print(min_idx)

        res = float('inf')
        for i in range(n-1, 2*n):
            if min_idx[i] - max_idx[i] < res:
                res = min_idx[i] - max_idx[i]
        
        return res

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