You are given a 0-indexed integer array nums
. Rearrange the values of nums
according to the following rules:
- Sort the values at odd indices of
nums
in non-increasing order.- For example, if
nums = [4,**1**,2,**3**]
before this step, it becomes[4,**3**,2,**1**]
after. The values at odd indices1
and3
are sorted in non-increasing order.
- For example, if
- Sort the values at even indices of
nums
in non-decreasing order.- For example, if
nums = [**4**,1,**2**,3]
before this step, it becomes[**2**,1,**4**,3]
after. The values at even indices0
and2
are sorted in non-decreasing order.
- For example, if
Return the array formed after rearranging the values of nums
.
Example 1:
Input: nums = [4,1,2,3]
Output: [2,3,4,1]
Explanation:
First, we sort the values present at odd indices (1 and 3) in non-increasing order.
So, nums changes from [4,1,2,3] to [4,3,2,1].
Next, we sort the values present at even indices (0 and 2) in non-decreasing order.
So, nums changes from [4,1,2,3] to [2,3,4,1].
Thus, the array formed after rearranging the values is [2,3,4,1].
Example 2:
Input: nums = [2,1]
Output: [2,1]
Explanation:
Since there is exactly one odd index and one even index, no rearrangement of values takes place.
The resultant array formed is [2,1], which is the same as the initial array.
Constraints:
-
1 <= nums.length <= 100
-
1 <= nums[i] <= 100
SOLUTION:
import heapq
class Solution:
def sortEvenOdd(self, nums: List[int]) -> List[int]:
n = len(nums)
evens = []
odds = []
for i in range(n):
if i & 1:
heapq.heappush(odds, -nums[i])
else:
heapq.heappush(evens, nums[i])
for i in range(n):
if i & 1:
nums[i] = -heapq.heappop(odds)
else:
nums[i] = heapq.heappop(evens)
return nums
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