#1968

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Array With Elements Not Equal to Average of Neighbors

Rearrange an array such that no element equals the average of its neighbors using a greedy approach.

Problem Statement

You are given a 0-indexed array nums of distinct integers. You need to rearrange the elements in the array so that every element is not equal to the average of its neighbors.

More formally, for every index i in the range 1 <= i < nums.length - 1, the condition (nums[i-1] + nums[i+1]) / 2 != nums[i] must hold. Return any rearrangement that meets these conditions.

Examples

Example 1

Input: nums = [1,2,3,4,5]

Output: [1,2,4,5,3]

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

Input: nums = [6,2,0,9,7]

Output: [9,7,6,2,0]

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. Note that the original array [6,2,0,9,7] also satisfies the conditions.

Constraints

3 <= nums.length <= 105
0 <= nums[i] <= 105

Solution Approach

Greedy Rearrangement

A direct greedy approach can solve this by sorting the array first and then swapping adjacent elements. This ensures that no element will be the average of its neighbors. By sorting the array, we minimize the risk of an element becoming the average when it's placed between smaller and larger numbers.

Invariant Validation

After sorting, the key invariant is that the swapped elements will prevent any number from becoming the average of its neighbors. By checking the neighbors after every rearrangement, we can validate that the condition is satisfied. The greedy approach guarantees that after proper swapping, the invariant holds for all indices.

Optimal Rearrangement Strategy

To efficiently rearrange the elements, consider a strategy where you place the smallest and largest numbers alternatively in the array. This approach ensures that the average condition is broken by keeping each element at least one step away from the average of its neighbors.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n log n) due to the sorting step. Rearranging the elements after sorting takes O(n), making the overall time complexity dominated by the sorting step. The space complexity is O(1) if the rearrangement is done in-place, or O(n) if a new array is used.

What Interviewers Usually Probe

Candidates should focus on the greedy nature of the solution, ensuring no element becomes the average of its neighbors.
Look for an understanding of invariants and how the rearranged array maintains these properties after sorting.
Evaluate the candidate's ability to handle edge cases where the smallest and largest values are at the boundaries of the array.

Common Pitfalls or Variants

Common pitfalls

Not considering the possibility of adjacent elements being too close, leading to violations of the average condition.
Failing to rearrange the array optimally, resulting in a non-working solution that does not break the neighbor average condition.
Overcomplicating the problem with unnecessary checks or operations that increase the time complexity.

Follow-up variants

Try the problem with arrays of varying lengths, from the minimum size to the maximum allowed by the constraints.
Test with arrays where values are arranged in increasing or decreasing order, and analyze the effect on the solution's correctness.
Introduce duplicate elements into the problem and adjust the constraints to see if the greedy approach can still be applied.

FAQ

What is the key to solving 'Array With Elements Not Equal to Average of Neighbors'?

The key is using a greedy approach where you rearrange the array such that no element equals the average of its neighbors. Sorting and swapping adjacent elements often solves this problem.

Can the input array have duplicate elements?

No, the problem specifies that the array contains distinct integers. If duplicates were allowed, the solution would need additional handling.

What is the time complexity of solving this problem?

The time complexity is O(n log n) due to the sorting step, followed by a linear pass to rearrange the elements, making the overall complexity O(n log n).

Is it possible to solve the problem without sorting the array?

While sorting is a common approach, the problem can be solved without sorting by directly rearranging the elements based on their relationship with their neighbors, though this could be less efficient.

How does GhostInterview help with the greedy approach for this problem?

GhostInterview offers step-by-step guidance through the greedy approach, ensuring you understand the reasoning behind each step and how to rearrange the array efficiently.

terminal

Solution

Solution 1: Sorting

Since the elements in the array are distinct, we can first sort the array, then divide the array into two parts. Place the first half of the elements in the even positions of the answer array, and the second half of the elements in the odd positions of the answer array. In this way, for each element, its two adjacent elements will not be equal to its average value.

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class Solution:
    def rearrangeArray(self, nums: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        m = (n + 1) // 2
        ans = []
        for i in range(m):
            ans.append(nums[i])
            if i + m < n:
                ans.append(nums[i + m])
        return ans

Continue Practicing

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