Leetcode #1982: Find Array Given Subset Sums
In this guide, we solve Leetcode #1982 Find Array Given Subset Sums in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer n representing the length of an unknown array that you are trying to recover. You are also given an array sums containing the values of all 2n subset sums of the unknown array (in no particular order).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Divide and Conquer
Intuition
The problem splits naturally into smaller subproblems.
Solving each part independently makes the whole problem manageable.
Approach
Divide the input, solve recursively, then merge results.
This structure often yields clean and efficient code.
Steps:
- Divide into subproblems.
- Solve recursively.
- Merge to get final answer.
Example
Input: n = 3, sums = [-3,-2,-1,0,0,1,2,3]
Output: [1,2,-3]
Explanation: [1,2,-3] is able to achieve the given subset sums:
- []: sum is 0
- [1]: sum is 1
- [2]: sum is 2
- [1,2]: sum is 3
- [-3]: sum is -3
- [1,-3]: sum is -2
- [2,-3]: sum is -1
- [1,2,-3]: sum is 0
Note that any permutation of [1,2,-3] and also any permutation of [-1,-2,3] will also be accepted.
Python Solution
class Solution:
def recoverArray(self, n: int, sums: List[int]) -> List[int]:
m = -min(sums)
sl = SortedList(x + m for x in sums)
sl.remove(0)
ans = [sl[0]]
for i in range(1, n):
for j in range(1 << i):
if j >> (i - 1) & 1:
s = sum(ans[k] for k in range(i) if j >> k & 1)
sl.remove(s)
ans.append(sl[0])
for i in range(1 << n):
s = sum(ans[j] for j in range(n) if i >> j & 1)
if s == m:
for j in range(n):
if i >> j & 1:
ans[j] *= -1
break
return ans
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(log n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.