You are given a 0-indexed integer array nums consisting of 3 * n elements.
You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:
n elements belonging to the first part and their sum is sumfirst.n elements belonging to the second part and their sum is sumsecond.The difference in sums of the two parts is denoted as sumfirst - sumsecond.
sumfirst = 3 and sumsecond = 2, their difference is 1.sumfirst = 2 and sumsecond = 3, their difference is -1.Return the minimum difference possible between the sums of the two parts after the removal of n elements.
\n
Example 1:
\n\n\nInput: nums = [3,1,2]\nOutput: -1\nExplanation: Here, nums has 3 elements, so n = 1. \nThus we have to remove 1 element from nums and divide the array into two equal parts.\n- 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.\n- 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.\n- 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.\nThe minimum difference between sums of the two parts is min(-1,1,2) = -1. \n\n\n
Example 2:
\n\n\nInput: nums = [7,9,5,8,1,3]\nOutput: 1\nExplanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.\nIf 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.\nTo 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.\nIt can be shown that it is not possible to obtain a difference smaller than 1.\n\n\n
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Constraints:
\n\nnums.length == 3 * n1 <= n <= 1051 <= nums[i] <= 105给你一个下标从 0 开始的整数数组 nums ，它包含 3 * n 个元素。
你可以从 nums 中删除 恰好 n 个元素，剩下的 2 * n 个元素将会被分成两个 相同大小 的部分。
n 个元素属于第一部分，它们的和记为 sumfirst 。n 个元素属于第二部分，它们的和记为 sumsecond 。两部分和的 差值 记为 sumfirst - sumsecond 。
sumfirst = 3 且 sumsecond = 2 ，它们的差值为 1 。sumfirst = 2 且 sumsecond = 3 ，它们的差值为 -1 。请你返回删除 n 个元素之后，剩下两部分和的 差值的最小值 是多少。
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示例 1：
\n\n输入：nums = [3,1,2]\n输出：-1\n解释：nums 有 3 个元素，所以 n = 1 。\n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。\n- 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 。\n- 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 。\n- 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 。\n两部分和的最小差值为 min(-1,1,2) = -1 。\n\n\n
示例 2：
\n\n输入：nums = [7,9,5,8,1,3]\n输出：1\n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。\n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 。\n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 。\n观察可知，最优答案为 1 。\n\n\n
\n\n
提示：
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