How do you solve Closest Subsequence Sum on LeetCode?

Closest Subsequence Sum (LeetCode #1755) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^(n/2)) time complexity and O(2^(n/2)) space complexity. Key insight: Dividing the problem into smaller parts can significantly reduce complexity.

What is the brute force solution for Closest Subsequence Sum?

The brute force approach for Closest Subsequence Sum is Brute Force, with O(2^n) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and calculating their sums. This method is straightforward but inefficient due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n * 2^(n/2)).

What algorithm or data structure is used to solve Closest Subsequence Sum?

Closest Subsequence Sum uses the following concepts: Array, Two Pointers, Dynamic Programming, Bit Manipulation, Sorting, Bitmask. The recommended patterns to study are: Meet-in-the-Middle, Binary Search, Backtracking.

What are the interview tips for Closest Subsequence Sum?

Always discuss your thought process and approach before coding. Be prepared to explain why you chose a particular algorithm or data structure. Practice breaking down problems into smaller parts to simplify your approach.

What are common mistakes when solving Closest Subsequence Sum?

Not considering all possible subsequences, leading to incorrect results. Failing to optimize the search for closest sums, resulting in excessive time complexity.

#1755

Closest Subsequence Sum

Hard

ArrayTwo PointersDynamic ProgrammingBit ManipulationSortingBitmaskMeet-in-the-MiddleBinary SearchBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(n * 2^(n/2))
Space	O(1)	O(2^(n/2))

💡

Intuition

Time O(n * 2^(n/2))Space O(2^(n/2))

The optimal solution leverages the idea of dividing the array into two halves and using a meet-in-the-middle approach. This reduces the problem size significantly and allows us to efficiently find the closest sum to the goal.

⚙️

Algorithm

4 steps

1Step 1: Split the array into two halves.
2Step 2: Generate all possible sums for both halves.
3Step 3: Sort one of the halves' sums to enable binary search.
4Step 4: For each sum from the first half, use binary search to find the closest sum in the second half that, when added, is closest to the goal.

solution.py34 lines

1# Full working Python code
2from itertools import combinations
3
4
5def closest_subsequence_sum(nums, goal):
6    n = len(nums)
7    left_part = nums[:n//2]
8    right_part = nums[n//2:]
9    left_sums = set()
10    right_sums = set()
11
12    for i in range(len(left_part) + 1):
13        for comb in combinations(left_part, i):
14            left_sums.add(sum(comb))
15
16    for i in range(len(right_part) + 1):
17        for comb in combinations(right_part, i):
18            right_sums.add(sum(comb))
19
20    left_sums = sorted(left_sums)
21    min_diff = float('inf')
22
23    for s in right_sums:
24        target = goal - s
25        l, r = 0, len(left_sums) - 1
26        while l <= r:
27            mid = (l + r) // 2
28            min_diff = min(min_diff, abs(left_sums[mid] + s - goal))
29            if left_sums[mid] + s < goal:
30                l = mid + 1
31            else:
32                r = mid - 1
33
34    return min_diff

ℹ

Complexity note: The time complexity is O(n * 2^(n/2)) because we generate all possible sums for two halves of the array (2^(n/2) each) and sort one of them. The space complexity is O(2^(n/2)) due to storing the sums.

1Dividing the problem into smaller parts can significantly reduce complexity.
2Using binary search on sorted sums allows for efficient closest sum calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.