LeetCode 2007. Find Original Array From Doubled Array

原题链接在这里：https://leetcode.com/problems/find-original-array-from-doubled-array/

题目：

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

题解：

Count the freq of each element.

Starting from smallest x, check if there are more 2x in the map.

To start from the smallest, we could use TreeMap.

Note: for(int j = 0; j < tm.get(x); j++) needs to be j < tm.get(x). tm.get(x) could be changed. e.g. changed = {0, 0, 0, 0}. the original is {0, 0}. tm.get(0) keeps changing in each loop iteration.

Time Complexity: O(nlogn). n = changed.length. 

Space: O(n).

AC Java:

 1 class Solution {
 2     public int[] findOriginalArray(int[] changed) {
 3         if(changed == null || changed.length % 2 == 1){
 4             return new int[0];
 5         }
 6         
 7         int n = changed.length;
 8         int [] res = new int[n / 2];
 9         TreeMap<Integer, Integer> tm = new TreeMap<>();
10         for(int num : changed){
11             tm.put(num, tm.getOrDefault(num, 0) + 1);
12         }
13         
14         int i = 0;
15         for(int x : tm.keySet()){
16             if(tm.get(x) > tm.getOrDefault(x + x, 0)){
17                 return new int[0];
18             }
19             
20             for(int j = 0; j < tm.get(x); j++){
21                 res[i++] = x;
22                 tm.put(x + x, tm.get(x + x) - 1);
23             }
24         }
25         
26         return res;
27     }
28 }