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Find Original Array From Doubled Array

Problem Description​

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. If changed is not a doubled array, return an empty array. The elements in original may be returned in any order.

Examples​

Example 1:

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].

Example 2:

Input: changed = [6,3,0,1]
Output: []
Explanation: changed is not a doubled array.

Example 3:

Input: changed = [1]
Output: []
Explanation: changed is not a doubled array.

Constraints​

  • 1 <= changed.length <= 10^5
  • 0 <= changed[i] <= 10^5

Approach​

To solve this problem efficiently:

  1. Use a hash map (freq) to count occurrences of each number in changed.
  2. Sort changed to process elements in increasing order.
  3. Iterate through each number num in sorted changed:
    • Check if freq[num] is greater than freq[2 * num]:
      • If yes, return an empty array since changed cannot be a doubled array.
    • Decrease freq[num] and freq[2 * num] accordingly.
  4. After iterating through all elements, construct and return the original array from freq.

This approach ensures correctness by leveraging sorting and counting with a hash map to validate if changed can indeed represent a doubled array of some original array original.

C++ Solution​

class Solution {
public:
    vector<int> findOriginalArray(vector<int>& changed) {
        unordered_map<int, int> freq;
        vector<int> original;
        
        for (int num : changed) {
            freq[num]++;
        }
        
        sort(changed.begin(), changed.end());
        
        for (int num : changed) {
            if (freq[num] > freq[2 * num]) {
                return {};
            }
            for (int i = 0; i < freq[num]; ++i) {
                original.push_back(num);
                freq[2 * num]--;
            }
            freq[num] = 0;
        }
        
        return original;
    }
};