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
\n
Constraints:
\n\nnums.length == 3 * n1 <= n <= 1051 <= nums[i] <= 105Compiled with clang 11 using the latest C++ 17 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
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