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Formatted question description: https://leetcode.ca/all/2164.html
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
Similar Questions:
Solution 1.
// OJ: https://leetcode.com/problems/sort-even-and-odd-indices-independently/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
vector<int> sortEvenOdd(vector<int>& A) {
vector<int> odd, even, ans;
for (int i = 0; i < A.size(); ++i) {
if (i % 2) odd.push_back(A[i]);
else even.push_back(A[i]);
}
sort(begin(odd), end(odd), greater<>());
sort(begin(even), end(even));
for (int i = 0, j = 0, k = 0; i < A.size(); ++i) {
if (i % 2) ans.push_back(odd[j++]);
else ans.push_back(even[k++]);
}
return ans;
}
};