## Array With Elements Not Equal to Average of Neighbors solution leetcode

You are given a **0-indexed** array `nums`

of **distinct** integers. You want to rearrange the elements in the array such that every element in the rearranged array is **not** equal to the **average** of its neighbors.

More formally, the rearranged array should have the property such that for every `i`

in the range `1 <= i < nums.length - 1`

, `(nums[i-1] + nums[i+1]) / 2`

is **not** equal to `nums[i]`

.

Return **any** rearrangement of `nums`

* that meets the requirements*.

**Example 1: Array With Elements Not Equal to Average of Neighbors solution leetcode**

Input:nums = [1,2,3,4,5]Output:[1,2,4,5,3]Explanation:When i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5. When i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5. When i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.

**Example 2: Array With Elements Not Equal to Average of Neighbors solution leetcode**

Input:nums = [6,2,0,9,7]Output:[9,7,6,2,0]Explanation:When i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5. When i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5. When i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.

**Constraints: Array With Elements Not Equal to Average of Neighbors solution leetcode**

`3 <= nums.length <= 10`

^{5}`0 <= nums[i] <= 10`

^{5}