Leetcode #2007: Find Original Array From Doubled Array

In this guide, we solve Leetcode #2007 Find Original Array From Doubled 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.

Leetcode

Problem Statement

An integer array original is transformed into a doubled array changed by appending twice the value of every element in original, and then randomly shuffling the resulting array. Given an array changed, return original if changed is a doubled array.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Greedy, Array, Hash Table, 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: changed = [1,3,4,2,6,8]
Output: [1,3,4]
Explanation: One possible original array could be [1,3,4]:
- Twice the value of 1 is 1 * 2 = 2.
- Twice the value of 3 is 3 * 2 = 6.
- Twice the value of 4 is 4 * 2 = 8.
Other original arrays could be [4,3,1] or [3,1,4].

Python Solution

class Solution:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        changed.sort()
        cnt = Counter(changed)
        ans = []
        for x in changed:
            if cnt[x] == 0:
                continue
            cnt[x] -= 1
            if cnt[x << 1] <= 0:
                return []
            cnt[x << 1] -= 1
            ans.append(x)
        return ans

Complexity

The time complexity is $O(n \times \log n)$ , and the space complexity is $O(n)$ , where $n$ is the length of the array changed. The space complexity is $O(n)$ , where $n$ is the length of the array changed.

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.

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.

Python Solution

class Solution:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        changed.sort()
        cnt = Counter(changed)
        ans = []
        for x in changed:
            if cnt[x] == 0:
                continue
            cnt[x] -= 1
            if cnt[x << 1] <= 0:
                return []
            cnt[x << 1] -= 1
            ans.append(x)
        return ans

Leetcode #2007: Find Original Array From Doubled Array

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2007: Find Original Array From Doubled Array

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview