Leetcode 1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

Input: The input consists of an array of integers `nums` and an integer `limit`.

Example: nums = [6,3,5,8], limit = 3

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^9

• 0 <= limit <= 10^9

Output: The output is an integer representing the size of the longest subarray satisfying the given condition.

Example: 2

Constraints:

• The output is a non-negative integer.

Goal: To calculate the size of the longest subarray where the absolute difference between the minimum and maximum elements is less than or equal to the given `limit`.

Steps:

• Step 1: Use a multiset to track the current subarray's elements.

• Step 2: For every element in `nums`, add it to the multiset and check if the max difference exceeds the `limit`.

• Step 3: If the condition is violated, remove elements from the start of the current subarray until it satisfies the condition again.

• Step 4: Keep track of the maximum subarray size.

Goal: Ensure the solution is efficient for large arrays and large values of `nums[i]` and `limit`.

Steps:

• Use a sliding window technique to maintain efficiency.

• Handle edge cases with a limit of 0.

Assumptions:

• The array `nums` contains at least one element.

• The input values are within the given constraints.

• Input: nums = [6,3,5,8], limit = 3

• Explanation: The longest subarray is [6,3] or [5,8], each with a size of 2.

• Input: nums = [9,2,4,6], limit = 4

• Explanation: The subarray [2,4,6] has a max difference of 4 and is the longest subarray with size 3.

Approach: Use a sliding window approach with a multiset to efficiently maintain the current subarray's minimum and maximum values.

Observations:

• The problem requires maintaining a subarray that satisfies a condition on the maximum and minimum values.

• A sliding window with dynamic size can help efficiently process the array.

• We can use a multiset or two deques to maintain the max and min values of the current window.

Steps:

• Step 1: Initialize a multiset or two deques for tracking min and max values.

• Step 2: Iterate through the array and add elements to the current subarray.

• Step 3: Remove elements from the front of the subarray if the condition is violated.

• Step 4: Track the size of the longest valid subarray.

Empty Inputs:

• No empty input is allowed as `nums` will always have at least one element.

Large Inputs:

• The solution must handle arrays with up to 10^5 elements efficiently.

Special Values:

• Handle cases with `limit = 0` where only subarrays with identical elements are valid.

Constraints:

• Ensure the solution remains efficient even for extreme values of `limit` or `nums[i]`.

int longestSubarray(vector<int>& nums, int limit) {
    multiset<int> ms;
    int res = 0, j = 0;
    for(int i = 0; i < nums.size(); i++) {
        ms.insert(nums[i]);
        while(!ms.empty() && *ms.rbegin() - *ms.begin() > limit) {
            ms.erase(ms.find(nums[j++]));
        }
        res = max(res, i - j + 1);
    }
    return res;
}

1 : Function Definition

int longestSubarray(vector<int>& nums, int limit) {

Defines the function `longestSubarray`, which takes a vector of integers `nums` and an integer `limit` as input. It returns the length of the longest subarray where the difference between the maximum and minimum elements is within the limit.

2 : Variable Initialization

    multiset<int> ms;

Initializes a `multiset` named `ms` to store elements of the subarray in sorted order. This allows efficient access to the minimum and maximum elements.

3 : Variable Initialization

    int res = 0, j = 0;

Initializes two variables: `res` to store the length of the longest valid subarray found and `j` to represent the starting index of the sliding window.

4 : Loop

    for(int i = 0; i < nums.size(); i++) {

Starts a loop where `i` iterates through each element of the `nums` vector. This loop represents the end of the sliding window.

5 : Insert Operation

        ms.insert(nums[i]);

Inserts the current element `nums[i]` into the `multiset` to maintain the sorted order of the subarray.

6 : Condition Check

        while(!ms.empty() && *ms.rbegin() - *ms.begin() > limit) {

Checks whether the difference between the maximum (`*ms.rbegin()`) and minimum (`*ms.begin()`) elements of the `multiset` exceeds the `limit`. If so, the subarray is invalid, and the sliding window is adjusted.

7 : Erase Operation

            ms.erase(ms.find(nums[j++]));

Erases the element at index `j` from the `multiset` and increments `j` to shrink the sliding window from the left side.

8 : Update Result

        res = max(res, i - j + 1);

Updates the result `res` with the length of the current valid subarray, which is `i - j + 1`, and stores the maximum length found.

9 : Return Statement

    return res;

Returns the value of `res`, which represents the length of the longest valid subarray found.

Best Case: O(n) where n is the length of `nums`.

Average Case: O(n) for maintaining the sliding window and set operations.

Worst Case: O(n) for iterating through the array.

Description: The time complexity is linear as each element is added and removed from the sliding window at most once.

Best Case: O(1) if subarrays are very small.

Worst Case: O(n) for the multiset or deque in extreme cases.

Description: The space complexity depends on the size of the sliding window and is O(n) in the worst case.

Leetcode 1438: Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

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