Leetcode 1968: Array With Elements Not Equal to Average of Neighbors
You are given an array
nums containing distinct integers. You need to rearrange the elements of the array such that no element is equal to the average of its two adjacent elements. Specifically, for every index i (1 <= i < nums.length - 1), the condition (nums[i-1] + nums[i+1]) / 2 != nums[i] should hold. Your task is to return any valid rearrangement of nums that satisfies this condition.Problem
Approach
Steps
Complexity
Input: The input is an array `nums` containing distinct integers.
Example: nums = [1, 2, 3, 4, 5]
Constraints:
• 3 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^5
• nums contains distinct integers
Output: The output is a rearranged array where no element is the average of its neighbors.
Example: Output: [1, 2, 4, 5, 3]
Constraints:
• The output should be a valid rearrangement of the input array.
Goal: The goal is to rearrange the array such that no element is the average of its adjacent elements.
Steps:
• Step 1: Sort the input array.
• Step 2: Split the sorted array into two halves: one with smaller elements and the other with larger elements.
• Step 3: Interleave elements from the two halves to ensure that no element is the average of its neighbors.
Goal: The constraints ensure that the solution is efficient and works within the input limits.
Steps:
• The input array contains distinct integers.
• The array length is between 3 and 10^5.
Assumptions:
• The input array always contains distinct integers.
• Input: Input: nums = [1, 2, 3, 4, 5], Output: [1, 2, 4, 5, 3]
• Explanation: In this example, we rearrange the elements such that no element is equal to the average of its neighbors. For instance, when i=2, nums[i]=4, and the average of its neighbors is (2+5)/2=3.5, which is not equal to nums[i].
• Input: Input: nums = [6, 2, 0, 9, 7], Output: [9, 7, 6, 2, 0]
• Explanation: After rearranging the array, we ensure that no element is the average of its neighbors. For example, when i=2, nums[i]=6, and the average of its neighbors is (7+2)/2=4.5, which is not equal to nums[i].
Approach: The solution involves sorting the array, splitting it into two parts, and interleaving elements from each part to achieve the desired rearrangement.
Observations:
• The key idea is to avoid placing an element in the middle of two elements where it could be the average of its neighbors.
• By splitting the array and interleaving the elements, we can ensure that no element is the average of its adjacent ones.
Steps:
• Step 1: Sort the array `nums` to ensure elements are in increasing order.
• Step 2: Split the sorted array into two halves, the first half being smaller elements and the second half larger elements.
• Step 3: Merge the two halves by alternating elements from the two halves, ensuring no element is the average of its neighbors.
Empty Inputs:
• The input will always have at least three elements, as per the problem constraints.
Large Inputs:
• The solution should efficiently handle input arrays of up to 10^5 elements.
Special Values:
• The input contains distinct integers, so there will be no duplicates to handle.
Constraints:
• The array length will always be between 3 and 10^5, and elements are distinct.
vector<int> rearrangeArray(vector<int>& nums) {
int n = nums.size();
auto midptr = nums.begin() + n / 2;
nth_element(nums.begin(), midptr, nums.end());
int mid = *midptr;
#define A(i) nums[(1 + 2 * i) % (n|1)]
int i = 0, j = 0, k = n - 1;
while(j <= k) {
if(A(j) < mid)
swap(A(j), A(k--));
else if(A(j) > mid)
swap(A(i++), A(j++));
else j++;
}
return nums;
}
1 : Function Definition
vector<int> rearrangeArray(vector<int>& nums) {
The function `rearrangeArray` is defined, which takes a vector of integers `nums` and rearranges it around its median.
2 : Array Size Calculation
int n = nums.size();
The size of the input vector `nums` is stored in the variable `n`.
3 : Pointer Calculation
auto midptr = nums.begin() + n / 2;
A pointer `midptr` is initialized to the middle element of the array.
4 : Find Median Using nth_element
nth_element(nums.begin(), midptr, nums.end());
The `nth_element` function is used to partition the array such that the element at `midptr` is the median of the array.
5 : Extract Median Value
int mid = *midptr;
The median value is extracted from the pointer `midptr` and stored in the variable `mid`.
6 : Macro Definition
#define A(i) nums[(1 + 2 * i) % (n|1)]
A macro `A(i)` is defined, which calculates the position of an element in the rearranged array using a circular indexing pattern.
7 : Variable Initialization
int i = 0, j = 0, k = n - 1;
The variables `i`, `j`, and `k` are initialized. `i` and `j` are used to track the current positions in the array, while `k` tracks the rightmost position.
8 : Loop Start
while(j <= k) {
A while loop starts, which continues until the `j` pointer exceeds `k`.
9 : Check Value Less Than Median
if(A(j) < mid)
If the element at position `A(j)` is less than the median, the element is swapped with the element at position `A(k)`.
10 : Swap Less Than Median
swap(A(j), A(k--));
The elements at positions `A(j)` and `A(k)` are swapped, and `k` is decremented.
11 : Check Value Greater Than Median
else if(A(j) > mid)
If the element at position `A(j)` is greater than the median, the element is swapped with the element at position `A(i)`.
12 : Swap Greater Than Median
swap(A(i++), A(j++));
The elements at positions `A(i)` and `A(j)` are swapped, and both `i` and `j` are incremented.
13 : Move to Next Element
else j++;
If the element at position `A(j)` is equal to the median, `j` is incremented.
14 : Return Rearranged Array
return nums;
The rearranged array `nums` is returned.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting step, which is O(n log n). The rest of the steps (splitting and merging) take linear time, O(n).
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the input array and temporary variables during the sorting and merging process.
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