In a warehouse, there is a row of barcodes, where the ith 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: barcodes = [1,1,1,2,2,2]
Output: [2,1,2,1,2,1]
Example 2:
Input: barcodes = [1,1,1,1,2,2,3,3]
Output: [1,3,1,3,1,2,1,2]
Constraints:
-
1 <= barcodes.length <= 10000 -
1 <= barcodes[i] <= 10000
SOLUTION:
import bisect
class Solution:
def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
n = len(barcodes)
ctr = {}
for b in barcodes:
ctr[b] = ctr.get(b, 0) + 1
freq = sorted([(-v, k) for k, v in ctr.items()])
op = []
while len(op) < n:
i = 0
curr = freq[i]
if len(op) > 0:
while curr[1] == op[-1]:
i += 1
curr = freq[i]
c = curr[1]
else:
c = curr[1]
op.append(c)
del freq[bisect.bisect_left(freq, curr)]
if curr[0] + 1 < 0:
bisect.insort(freq, (curr[0] + 1, curr[1]))
return op
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