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Formatted question description: https://leetcode.ca/all/1755.html
1755. Closest Subsequence Sum
Level
Hard
Description
You are given an integer array nums
and an integer goal
.
You want to choose a subsequence of nums
such that the sum of its elements is the closest possible to goal
. That is, if the sum of the subsequence’s elements is sum
, then you want to minimize the absolute difference abs(sum  goal)
.
Return the minimum possible value of abs(sum  goal)
.
Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.
Example 1:
Input: nums = [5,7,3,5], goal = 6
Output: 0
Explanation: Choose the whole array as a subsequence, with a sum of 6. This is equal to the goal, so the absolute difference is 0.
Example 2:
Input: nums = [7,9,15,2], goal = 5
Output: 1
Explanation: Choose the subsequence [7,9,2], with a sum of 4.
The absolute difference is abs(4  (5)) = abs(1) = 1, which is the minimum.
Example 3:
Input: nums = [1,2,3], goal = 7
Output: 7
Constraints:
1 <= nums.length <= 40
10^7 <= nums[i] <= 10^7
10^9 <= goal <= 10^9
Solution
Divide nums
of two parts that have lengths differ by at most 1. For each part, calculate all subsets’ sums of elements, and use a list to store the sums of all subsets. After the two lists are obtained, sort the two lists and find the closest sum of two elements from the two list (one from each list). Use two pointers to do this. After the closest sum is found, calculate the minimum absolute difference.

class Solution { public int minAbsDifference(int[] nums, int goal) { if (goal == 0) return 0; Arrays.sort(nums); int length = nums.length; int halfLength = length / 2; List<Integer> list1 = new ArrayList<Integer>(); list1.add(0); List<Integer> list2 = new ArrayList<Integer>(); list2.add(0); for (int i = 0; i < halfLength; i++) { int num = nums[i]; int size = list1.size(); for (int j = 0; j < size; j++) list1.add(list1.get(j) + num); } for (int i = halfLength; i < length; i++) { int num = nums[i]; int size = list2.size(); for (int j = 0; j < size; j++) list2.add(list2.get(j) + num); } Collections.sort(list1); Collections.sort(list2); int minDifference = Integer.MAX_VALUE; int size1 = list1.size(), size2 = list2.size(); int pointer1 = 0, pointer2 = size2  1; while (pointer1 < size1 && pointer2 >= 0) { int curSum = list1.get(pointer1) + list2.get(pointer2); int curDifference = curSum  goal; minDifference = Math.min(minDifference, Math.abs(curDifference)); if (curDifference == 0) return 0; else if (curDifference < 0) pointer1++; else pointer2; } return minDifference; } }

Todo

class Solution: def minAbsDifference(self, nums: List[int], goal: int) > int: n = len(nums) left = set() right = set() self.getSubSeqSum(0, 0, nums[: n // 2], left) self.getSubSeqSum(0, 0, nums[n // 2 :], right) result = inf right = sorted(right) rl = len(right) for l in left: remaining = goal  l idx = bisect_left(right, remaining) if idx < rl: result = min(result, abs(remaining  right[idx])) if idx > 0: result = min(result, abs(remaining  right[idx  1])) return result def getSubSeqSum(self, i: int, curr: int, arr: List[int], result: Set[int]): if i == len(arr): result.add(curr) return self.getSubSeqSum(i + 1, curr, arr, result) self.getSubSeqSum(i + 1, curr + arr[i], arr, result)