Leetcode #2122: Recover the Original Array
In this guide, we solve Leetcode #2122 Recover the Original Array 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
Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary positive integer k and created two new 0-indexed integer arrays lower and higher in the following manner: lower[i] = arr[i] - k, for every index i where 0 <= i < n higher[i] = arr[i] + k, for every index i where 0 <= i < n Unfortunately, Alice lost all three arrays.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Hash Table, Two Pointers, Enumeration, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [2,10,6,4,8,12]
Output: [3,7,11]
Explanation:
If arr = [3,7,11] and k = 1, we get lower = [2,6,10] and higher = [4,8,12].
Combining lower and higher gives us [2,6,10,4,8,12], which is a permutation of nums.
Another valid possibility is that arr = [5,7,9] and k = 3. In that case, lower = [2,4,6] and higher = [8,10,12].
Python Solution
class Solution:
def recoverArray(self, nums: List[int]) -> List[int]:
nums.sort()
n = len(nums)
for i in range(1, n):
d = nums[i] - nums[0]
if d == 0 or d % 2 == 1:
continue
vis = [False] * n
vis[i] = True
ans = [(nums[0] + nums[i]) >> 1]
l, r = 1, i + 1
while r < n:
while l < n and vis[l]:
l += 1
while r < n and nums[r] - nums[l] < d:
r += 1
if r == n or nums[r] - nums[l] > d:
break
vis[r] = True
ans.append((nums[l] + nums[r]) >> 1)
l, r = l + 1, r + 1
if len(ans) == (n >> 1):
return ans
return []
Complexity
The time complexity is O(n). The space complexity is O(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.