-
Notifications
You must be signed in to change notification settings - Fork 613
/
1054.py
49 lines (36 loc) · 1.18 KB
/
1054.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
'''
In a warehouse, there is a row of barcodes, where the i-th barcode is barcodes[i].
Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.
Example 1:
Input: [1,1,1,2,2,2]
Output: [2,1,2,1,2,1]
Example 2:
Input: [1,1,1,1,2,2,3,3]
Output: [1,3,1,3,2,1,2,1]
Note:
1 <= barcodes.length <= 10000
1 <= barcodes[i] <= 10000
'''
class Solution(object):
def rearrangeBarcodes(self, barcodes):
"""
:type barcodes: List[int]
:rtype: List[int]
"""
import heapq
di = collections.Counter(barcodes)
pq = [(-value, key) for key, value in di.items()]
heapq.heapify(pq)
# print pq
result = []
while len(pq) >= 2:
freq1, barcode1 = heapq.heappop(pq)
freq2, barcode2 = heapq.heappop(pq)
result.extend([barcode1, barcode2])
if freq1 + 1:
heapq.heappush(pq, (freq1 + 1, barcode1))
if freq2 + 1:
heapq.heappush(pq, (freq2 + 1, barcode2))
if pq:
result.append(pq[0][1])
return result