2163. Minimum Difference in Sums After Removal of Elements
Hard
50 Points
Array
Dynamic Programming
Heap (Priority Queue)
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:
The difference in sums of the two parts is denoted as sumfirst - sumsecond.
Return the minimum difference possible between the sums of the two parts after the removal of n elements.
Examples
Example 1
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.
Example 2
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.
Constraints
nums.length == 3 * n
1 <= n <= 105
1 <= nums[i] <= 105
2163. Minimum Difference in Sums After Removal of Elements
Hard
50 Points
Array
Dynamic Programming
Heap (Priority Queue)
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:
The difference in sums of the two parts is denoted as sumfirst - sumsecond.
Return the minimum difference possible between the sums of the two parts after the removal of n elements.
Examples
Example 1
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.
Example 2
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.
Constraints
nums.length == 3 * n
1 <= n <= 105
1 <= nums[i] <= 105
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