Leetcode 2966: Divide Array Into Arrays With Max Difference
You are given an array
nums of size n where n is a multiple of 3, and a positive integer k. Your task is to divide the array into n / 3 subarrays, each containing exactly 3 elements, such that the difference between the largest and smallest element in each subarray is less than or equal to k. If it’s possible to divide the array in this way, return a 2D array containing the subarrays. If it is not possible, return an empty array. If multiple valid divisions exist, return any valid one.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` of size `n` and an integer `k`. The array is divisible by 3 to allow division into valid groups of three elements.
Example: nums = [1, 3, 4, 8, 7, 9, 3, 5, 1], k = 2
Constraints:
• n == nums.length
• 1 <= n <= 10^5
• n is a multiple of 3
• 1 <= nums[i] <= 10^5
• 1 <= k <= 10^5
Output: Return a 2D array containing the subarrays formed by grouping the elements, where each group contains exactly 3 elements, and the difference between the maximum and minimum element in each group is at most `k`. If this is not possible, return an empty array.
Example: [[1, 1, 3], [3, 4, 5], [7, 8, 9]]
Constraints:
Goal: The goal is to divide the array into subarrays of size 3 where the difference between the maximum and minimum integers in each group is less than or equal to `k`.
Steps:
• Sort the array `nums` in ascending order.
• Iterate through the sorted array and try to form subarrays of size 3.
• For each subarray, check if the difference between the largest and smallest element is less than or equal to `k`.
• If all subarrays are valid, return them; otherwise, return an empty array.
Goal: The size of the array is guaranteed to be divisible by 3. Each subarray must contain exactly three elements, and the difference between the largest and smallest element in each subarray must be at most `k`.
Steps:
• 1 <= n <= 10^5
• n is a multiple of 3
• 1 <= nums[i] <= 10^5
• 1 <= k <= 10^5
Assumptions:
• The array can always be divided into groups of three, as long as the division criteria are met.
• The order of the elements within each subarray can vary, as long as the condition on the difference between the max and min is satisfied.
• Input: Input: nums = [1, 3, 4, 8, 7, 9, 3, 5, 1], k = 2
• Explanation: The array is first sorted as [1, 1, 3, 3, 4, 5, 7, 8, 9]. The valid subarrays are [[1, 1, 3], [3, 4, 5], [7, 8, 9]]. The difference between the maximum and minimum elements in each subarray is at most 2, so the output is valid.
• Input: Input: nums = [2, 4, 2, 2, 5, 2], k = 2
• Explanation: It is impossible to create valid subarrays because one subarray will always have the elements 2 and 5, and 5 - 2 = 3, which is greater than `k = 2`.
Approach: Sort the array and check if the difference between the largest and smallest integers in each group is within the allowed limit `k`. If it is, return the groups; otherwise, return an empty array.
Observations:
• The array needs to be divided into groups of three integers, which implies that the length of the array is divisible by 3.
• Sorting the array helps ensure that we can check the condition on differences in an ordered manner.
• By sorting the array, we can easily check if any adjacent group of three integers satisfies the condition on the difference.
Steps:
• Sort the array `nums` in ascending order.
• Iterate over the sorted array in chunks of three elements.
• For each chunk, check if the difference between the first and last element is <= `k`.
• If all chunks are valid, return them as the result; otherwise, return an empty array.
Empty Inputs:
• The array is guaranteed not to be empty as per the constraints.
Large Inputs:
• Ensure that the solution efficiently handles the maximum array length (10^5).
Special Values:
• The value of `k` can be large, which may make it difficult to form valid groups, so ensure that this is handled correctly.
Constraints:
• Check if the array length is divisible by 3 and handle cases where it's not.
vector<vector<int>> divideArray(vector<int>& nums, int k) {
int size = nums.size();
if (size % 3 != 0)
return vector<vector<int>>();
sort(nums.begin(), nums.end());
vector<vector<int>> result(size / 3, vector<int>(3));
int groupIndex = 0;
for (int i = 0; i < size; i += 3) {
if (i + 2 < size && nums[i + 2] - nums[i] <= k) {
result[groupIndex] = { nums[i], nums[i + 1], nums[i + 2] };
groupIndex++;
}
else {
return vector<vector<int>>();
}
}
return result;
}
1 : Function Definition
vector<vector<int>> divideArray(vector<int>& nums, int k) {
Defines the function `divideArray`, which takes a reference to an array `nums` and an integer `k` as input, returning a 2D vector of integers.
2 : Variable Initialization
int size = nums.size();
Initializes a variable `size` to store the length of the input array `nums`.
3 : Edge Case Check
if (size % 3 != 0)
Checks if the size of the array is a multiple of 3. If not, the array cannot be divided into groups of three.
4 : Return Empty Case
return vector<vector<int>>();
Returns an empty 2D vector if the size of the array is not divisible by 3.
5 : Sorting
sort(nums.begin(), nums.end());
Sorts the input array `nums` in ascending order to facilitate grouping the numbers.
6 : Result Initialization
vector<vector<int>> result(size / 3, vector<int>(3));
Initializes the `result` 2D vector, which will hold the groups of three integers.
7 : Group Index Initialization
int groupIndex = 0;
Initializes a variable `groupIndex` to keep track of the current index in the `result` vector.
8 : Loop Through Elements
for (int i = 0; i < size; i += 3) {
Begins a loop that iterates through the sorted array, stepping by 3 elements each time to form groups.
9 : Group Validation
if (i + 2 < size && nums[i + 2] - nums[i] <= k) {
Checks if the group of three consecutive elements has a difference between the largest and smallest element less than or equal to `k`.
10 : Group Assignment
result[groupIndex] = { nums[i], nums[i + 1], nums[i + 2] };
Assigns the current group of three numbers to the `result` vector at the current `groupIndex`.
11 : Group Index Increment
groupIndex++;
Increments the `groupIndex` to move to the next position in the `result` vector for the next group.
12 : Return Empty Case
else {
Handles the case where the group of numbers cannot be formed because the difference between the largest and smallest number is greater than `k`.
13 : Return Empty
return vector<vector<int>>();
Returns an empty 2D vector if the group is invalid.
14 : Return Result
return result;
Returns the `result` vector containing all the valid groups formed from the array.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The dominant factor is the sorting operation, which takes O(n log n).
Best Case: O(n)
Worst Case: O(n)
Description: We need extra space to store the subarrays, which in the worst case is proportional to the size of the input.
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