Even Odd: NLogN-Time, N-Space
Easy-LC problem today: Sort Even and Odd Indices Independently - LeetCode
2164. Sort Even and Odd Indices Independently
Easy
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
Given the small constraints, we can afford to do a NLogN-Time and N-Space solution:
1/ Copy the even and odd elements in their own arrays
2/ Sort them independently
3/ Merge them back
Code is down below, cheers, ACC.
public int[] SortEvenOdd(int[] nums) { int[] even = new int[nums.Length % 2 == 0 ? nums.Length / 2 : (nums.Length / 2 + 1)]; int[] odd = new int[nums.Length / 2]; int indexEven = 0; int indexOdd = 0; for (int i = 0; i < nums.Length; i++) { if (i % 2 == 0) even[indexEven++] = nums[i]; else odd[indexOdd++] = nums[i]; } Array.Sort(even); Array.Sort(odd); int[] retVal = new int[nums.Length]; indexEven = 0; indexOdd = odd.Length - 1; for (int i = 0; i < retVal.Length; i++) { if (i % 2 == 0) retVal[i] = even[indexEven++]; else retVal[i] = odd[indexOdd--]; } return retVal; }
Nice and very helpful article keep posting such a great articles ...
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