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[9/15] 2007. Find Original Array From Doubled Array

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2007. Find Original Array From Doubled Array

Medium

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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

Accepted

90,411

Submissions

221,593

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링크

https://leetcode.com/problems/find-original-array-from-doubled-array/ 3846 회 연결

익명
작성일 2022.09.15 09:23

from collections import Counter
class Solution:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        if len(changed)%2: return []
        changed = sorted(changed)
        N = len(changed)
        
        zero_counter = 0
        while changed and changed[0] == 0:
            changed.pop(0)
            zero_counter+=1
        
        if zero_counter%2: return []
        
        og = [0]*(zero_counter//2)
        counter = Counter(changed)
        
        for key in sorted(counter.keys()):
            if counter[key] == 0: continue
            if counter[key*2] == 0: return []
            counter[key*2] -= counter[key]
            

        
        for key in counter.keys():
            og += [key]*counter[key]
        return og if len(og)*2 == N else []

Runtime: 2829 ms, faster than 26.21% of Python3 online submissions for Find Original Array From Doubled Array.
Memory Usage: 33.3 MB, less than 21.38% of Python3 online submissions for Find Original Array From Doubled Array.
[code=Python 3]from collections import Counter
class Solution:
    def findOriginalArray(self, changed: List[int]) -> List[int]:
        if len(changed)%2: return []
        changed = sorted(changed)
        N = len(changed)
        
        zero_counter = 0
        while changed and changed[0] == 0:
            changed.pop(0)
            zero_counter+=1
        
        if zero_counter%2: return []
        
        og = [0]*(zero_counter//2)
        counter = Counter(changed)
        
        for key in sorted(counter.keys()):
            if counter[key] == 0: continue
            if counter[key*2] == 0: return []
            counter[key*2] -= counter[key]

for key in counter.keys():
            og += [key]*counter[key]
        return og if len(og)*2 == N else [][/code]