2007. Find Original Array From Doubled Array| Leetcode solution

 2007. Find Original Array From Doubled Array

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.

 

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 <= 105
• 0 <= changed[i] <= 105

solution:

class Solution {

public:

    vector<int> findOriginalArray(vector<int>& changed) {

        vector<int>result;

        if(changed.size() %2 != 0)

            return {};

        unordered_map<int, int>a;

        for(int i =0 ; i < changed.size() ;i++)

        {

            a[changed[i]] += 1;

        }

        sort(changed.begin() , changed.end());

        for(int i = 0 ; i < changed.size() ; i++)

        {    

            if(changed[i] == 0 && a[changed[i]] %2 == 0 && a[changed[i]]>0)

            {

                result.push_back(changed[i]);

                a[changed[i]] -= 2;

            }

              else  

             if(a[changed[i]]> 0 && a[2*changed[i]] > 0 && changed[i] != 0)

             {

                 result.push_back(changed[i]);

                 a[changed[i]] -= 1;

                 a[2*changed[i]] -= 1;

             }

            

        }

        if(result.size() == changed.size()/2)

            return result;

        return {};

    }

};