What is the time complexity of Array With Elements Not Equal to Average of Neighbors?

The optimal solution for Array With Elements Not Equal to Average of Neighbors runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) for the result array.

What is the brute force solution for Array With Elements Not Equal to Average of Neighbors?

The brute force approach for Array With Elements Not Equal to Average of Neighbors is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach involves generating all possible permutations of the array and checking each one to see if it meets the condition. This is simple but inefficient, as the number of permutations grows factorially with the size of the array. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Array With Elements Not Equal to Average of Neighbors?

Array With Elements Not Equal to Average of Neighbors uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Array With Elements Not Equal to Average of Neighbors?

Always start with a simple approach to understand the problem better. Discuss your thought process and reasoning behind each step during the interview. Practice explaining your code clearly and concisely, as communication is key.

What are common mistakes when solving Array With Elements Not Equal to Average of Neighbors?

Relying on brute-force without considering more efficient methods. Not checking edge cases, such as arrays with only three elements.

#1968

Array With Elements Not Equal to Average of Neighbors

Medium

ArrayGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n log n)
Space	O(n)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach involves sorting the array and then rearranging it by placing smaller elements at odd indices and larger elements at even indices. This ensures that no element can be the average of its neighbors.

⚙️

Algorithm

3 steps

1Step 1: Sort the array.
2Step 2: Create a new array of the same length.
3Step 3: Fill the new array with elements from the sorted array, placing smaller elements at odd indices and larger elements at even indices.

solution.py11 lines

1def rearrange(nums):
2    nums.sort()
3    n = len(nums)
4    result = [0] * n
5    mid = (n + 1) // 2
6    result[::2] = nums[:mid]
7    result[1::2] = nums[mid:]
8    return result
9
10# Example usage
11print(rearrange([1, 2, 3, 4, 5]))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) for the result array.

1Elements can only be the average of their neighbors if one neighbor is smaller and the other is larger.
2By sorting and distributing elements based on their size, we can avoid the average condition.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.