I knew the general shape of the solution pretty quickly: sort by absolute value, use a counter, greedily match each element x with 2x and collect x as part of the answer.
Recognize that the original array elements are exactly half of the doubled values, so we need to pair each element with its double. Use a hash map to count frequencies, then for each element in sorted order, if its count is positive, pair it with its double, decrement counts, and add the element to the result. This greedy approach works because sorting ensures we process smaller elements first, avoiding conflicts.
Pro tip: Clarify that the original array can be in any order, so returning any valid permutation is acceptable. Also, mention that if the problem guarantees a solution, we don't need to handle invalid cases, but we can discuss how to detect them.
Restate the problem: given an array of even length where each element is double some original element, recover any valid original array. Note that the original array can be in any order, and the input is guaranteed to have at least one valid original array.
Use a hash map (dictionary) to count the occurrences of each number in the input array. This allows O(1) lookups and updates when pairing elements.
Sorting ensures that when we process a number, its double (if present) will be larger, so we won't accidentally use a number that should be paired with a smaller number. This greedy strategy is optimal.
For each number in sorted order, if its count is positive, check if its double exists with positive count. If so, decrement both counts, add the number to the result, and continue. If not, the input is invalid (though guaranteed valid).
After processing all elements, return the list of original numbers. Discuss time complexity: O(n log n) due to sorting, and space complexity: O(n) for the hash map and result.
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