2164 - Sort Even and Odd Indices Independently (Easy)
Problem Link
https://leetcode.com/problems/sort-even-and-odd-indices-independently/
Problem Statement
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
Approach 1: Iteration
The simplest approach is to store odd-indexed numbers and even-indexed numbers separately, then sort them and merge them. One approach to store them separately is to check the parity (odd or even) of the iterating variable ( i
in this example solution), and store them to their corresponding lists.
We then sort the two lists. We note that odd-indexed numbers need to be non-increasing, so we need to call reverse
. Another approach to sort in reverse order is odds.sort(reverse = True)
.
We then put the numbers back to nums
using a similar method of retrieving the parity of the iterating variable.
def sortEvenOdd(self, nums: List[int]) -> List[int]:
#initialize the two lists for odd-indexed numbers and even-indexed numbers
odds = []
evens = []
#place the numbers into the two lists accordingly
for i in range(len(nums)):
if(i % 2 == 1):
odds.append(nums[i])
else:
evens.append(nums[i])
#sort the two lists
odds.sort()
evens.sort()
#as we want odd-indexed numbers to be non-increasing, we need to reverse it
odds.reverse()
#initialize result
result = []
#iterate all numbers
for i in range(len(nums)):
#put the numbers back
#we have i//2 because we take turns visiting odds and evens
if(i % 2 == 1):
result.append(odds[i//2])
else:
result.append(evens[i//2])
#return result
return result