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Author: jimit105

2267-minimum-difference-in-sums-after-removal-of-elements/README.md

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+<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>

2267-minimum-difference-in-sums-after-removal-of-elements/solution.py

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+class Solution:
+    def minimumDifference(self, nums: List[int]) -> int:
+        n3, n = len(nums), len(nums) // 3
+
+        part1 = [0] * (n + 1)
+        # max heap
+        total = sum(nums[:n])
+        ql = [-x for x in nums[:n]]
+        heapq.heapify(ql)
+        part1[0] = total
+
+        for i in range(n, n * 2):
+            total += nums[i]
+            heapq.heappush(ql, -nums[i])
+            total -= -heapq.heappop(ql)
+            part1[i - (n - 1)] = total
+
+        # min heap
+        part2 = sum(nums[n * 2 :])
+        qr = nums[n * 2 :]
+        heapq.heapify(qr)
+        ans = part1[n] - part2
+
+        for i in range(n * 2 - 1, n - 1, -1):
+            part2 += nums[i]
+            heapq.heappush(qr, nums[i])
+            part2 -= heapq.heappop(qr)
+            ans = min(ans, part1[i - n] - part2)
+
+        return ans