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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:

  1. 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 indices 1 and 3 are sorted in non-increasing order.
  2. 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 indices 0 and 2 are sorted in non-decreasing order.

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.

  • class Solution {
    public int[] sortEvenOdd(int[] nums) {
        int n = nums.length;
        int[] a = new int[(n + 1) >> 1];
        int[] b = new int[n >> 1];
        for (int i = 0, j = 0; j<n> > 1; i += 2, ++j) {
            a[j] = nums[i];
            b[j] = nums[i + 1];
        }
        if (n % 2 == 1) {
            a[a.length - 1] = nums[n - 1];
        }
        Arrays.sort(a);
        Arrays.sort(b);
        int[] ans = new int[n];
        for (int i = 0, j = 0; j < a.length; i += 2, ++j) {
            ans[i] = a[j];
        }
        for (int i = 1, j = b.length - 1; j >= 0; i += 2, --j) {
            ans[i] = b[j];
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> sortEvenOdd(vector<int>& nums) {
        int n = nums.size();
        vector<int> a;
        vector<int> b;
        for (int i = 0; i < n; ++i) {
            if (i % 2 == 0)
                a.push_back(nums[i]);
            else
                b.push_back(nums[i]);
        }
        sort(a.begin(), a.end());
        sort(b.begin(), b.end(), greater<int>());
        vector<int> ans(n);
        for (int i = 0, j = 0; j < a.size(); i += 2, ++j) ans[i] = a[j];
        for (int i = 1, j = 0; j < b.size(); i += 2, ++j) ans[i] = b[j];
        return ans;
    }
};

  • class Solution:
    def sortEvenOdd(self, nums: List[int]) -> List[int]:
        a = sorted(nums[::2])
        b = sorted(nums[1::2], reverse=True)
        nums[::2] = a
        nums[1::2] = b
        return nums


  • func sortEvenOdd(nums []int) []int {
	n := len(nums)
	var a []int
	var b []int
	for i, v := range nums {
		if i%2 == 0 {
			a = append(a, v)
		} else {
			b = append(b, v)
		}
	}
	ans := make([]int, n)
	sort.Ints(a)
	sort.Slice(b, func(i, j int) bool {
		return b[i] > b[j]
	})
	for i, j := 0, 0; j < len(a); i, j = i+2, j+1 {
		ans[i] = a[j]
	}
	for i, j := 1, 0; j < len(b); i, j = i+2, j+1 {
		ans[i] = b[j]
	}
	return ans
}

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