minimum-difference-in-sums-after-removal-of-elements

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Algorithms

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

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

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

The difference in sums of the two parts is denoted as <code>sumfirst - sumsecond</code>.

<ul>
	<li>For example, if <code>sumfirst = 3</code> and <code>sumsecond = 2</code>, their difference is <code>1</code>.</li>
	<li>Similarly, if <code>sumfirst = 2</code> and <code>sumsecond = 3</code>, their difference is <code>-1</code>.</li>
</ul>

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

&nbsp;
Example 1:

<pre>
Input: nums = [3,1,2]
Output: -1
Explanation: 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>

Example 2:

<pre>
Input: nums = [7,9,5,8,1,3]
Output: 1
Explanation: 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>

&nbsp;
Constraints:

<ul>
	<li><code>nums.length == 3 * n</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 105</code></li>
</ul>

minimumDifference

Minimum Difference in Sums After Removal of Elements

Product of Array Except Self

product-of-array-except-self

除自身以外数组的乘积

Find Subsequence of Length K With the Largest Sum

find-subsequence-of-length-k-with-the-largest-sum

找到和最大的长度为 K 的子序列

Array

Dynamic Programming

Heap (Priority Queue)

给你一个下标从 0&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，它包含&nbsp;<code>3 * n</code>&nbsp;个元素。

你可以从 <code>nums</code>&nbsp;中删除 恰好&nbsp;<code>n</code>&nbsp;个元素，剩下的 <code>2 * n</code>&nbsp;个元素将会被分成两个 相同大小&nbsp;的部分。

<ul>
	<li>前面&nbsp;<code>n</code>&nbsp;个元素属于第一部分，它们的和记为&nbsp;<code>sumfirst</code>&nbsp;。</li>
	<li>后面&nbsp;<code>n</code>&nbsp;个元素属于第二部分，它们的和记为&nbsp;<code>sumsecond</code>&nbsp;。</li>
</ul>

两部分和的 差值&nbsp;记为&nbsp;<code>sumfirst - sumsecond</code>&nbsp;。

<ul>
	<li>比方说，<code>sumfirst = 3</code> 且&nbsp;<code>sumsecond = 2</code>&nbsp;，它们的差值为&nbsp;<code>1</code>&nbsp;。</li>
	<li>再比方，<code>sumfirst = 2</code> 且&nbsp;<code>sumsecond = 3</code>&nbsp;，它们的差值为&nbsp;<code>-1</code>&nbsp;。</li>
</ul>

请你返回删除 <code>n</code>&nbsp;个元素之后，剩下两部分和的 差值的最小值&nbsp;是多少。

&nbsp;

示例 1：

<pre>输入：nums = [3,1,2]
输出：-1
解释：nums 有 3 个元素，所以 n = 1 。
所以我们需要从 nums 中删除 1 个元素，并将剩下的元素分成两部分。
- 如果我们删除 nums[0] = 3 ，数组变为 [1,2] 。两部分和的差值为 1 - 2 = -1 。
- 如果我们删除 nums[1] = 1 ，数组变为 [3,2] 。两部分和的差值为 3 - 2 = 1 。
- 如果我们删除 nums[2] = 2 ，数组变为 [3,1] 。两部分和的差值为 3 - 1 = 2 。
两部分和的最小差值为 min(-1,1,2) = -1 。
</pre>

示例 2：

<pre>输入：nums = [7,9,5,8,1,3]
输出：1
解释：n = 2 。所以我们需要删除 2 个元素，并将剩下元素分为 2 部分。
如果我们删除元素 nums[2] = 5 和 nums[3] = 8 ，剩下元素为 [7,9,1,3] 。和的差值为 (7+9) - (1+3) = 12 。
为了得到最小差值，我们应该删除 nums[1] = 9 和 nums[4] = 1 ，剩下的元素为 [7,5,8,3] 。和的差值为 (7+5) - (8+3) = 1 。
观察可知，最优答案为 1 。
</pre>

&nbsp;

提示：

<ul>
	<li><code>nums.length == 3 * n</code></li>
	<li><code>1 &lt;= n &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 105</code></li>
</ul>