What is the time complexity of Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

The optimal solution for Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit runs in O(n) time and uses O(n) space. The time complexity is O(n) because each element is processed at most twice (once added and once removed). The space complexity is O(n) due to the deques storing indices of the elements.

What is the brute force solution for Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

The brute force approach for Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible subarray and calculate the maximum and minimum values to see if their difference is within the limit. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit uses the following concepts: Array, Queue, Sliding Window, Heap (Priority Queue), Ordered Set, Monotonic Queue. The recommended patterns to study are: Sliding Window, Two Pointers, Deque.

What are the interview tips for Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

Always consider edge cases, such as when the limit is zero. Practice explaining your thought process as you code, as it helps clarify your understanding. Be prepared to discuss time and space complexity for your solution.

What are common mistakes when solving Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit?

Not using a sliding window approach, leading to unnecessary recalculations. Failing to update the left pointer correctly when the limit is exceeded.

#1438

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

Medium

ArrayQueueSliding WindowHeap (Priority Queue)Ordered SetMonotonic QueueSliding WindowTwo PointersDeque

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

We can use a sliding window approach to efficiently find the longest subarray. By maintaining the maximum and minimum values in the current window, we can quickly check if the absolute difference is within the limit.

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Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) and a variable to track the maximum length.
2Step 2: Use a data structure (like a deque) to maintain the maximum and minimum values in the current window.
3Step 3: Expand the right pointer to include more elements until the condition (max - min > limit) is violated.
4Step 4: If violated, move the left pointer to shrink the window until the condition is satisfied again.
5Step 5: Update the maximum length whenever the current window is valid.

solution.py22 lines

1from collections import deque
2
3def longestSubarray(nums, limit):
4    left = 0
5    max_length = 0
6    max_deque = deque()
7    min_deque = deque()
8    for right in range(len(nums)):
9        while max_deque and nums[max_deque[-1]] <= nums[right]:
10            max_deque.pop()
11        max_deque.append(right)
12        while min_deque and nums[min_deque[-1]] >= nums[right]:
13            min_deque.pop()
14        min_deque.append(right)
15        while nums[max_deque[0]] - nums[min_deque[0]] > limit:
16            left += 1
17            if max_deque[0] < left:
18                max_deque.popleft()
19            if min_deque[0] < left:
20                min_deque.popleft()
21        max_length = max(max_length, right - left + 1)
22    return max_length

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Complexity note: The time complexity is O(n) because each element is processed at most twice (once added and once removed). The space complexity is O(n) due to the deques storing indices of the elements.

1Using a sliding window allows us to efficiently manage the current subarray without recalculating max/min from scratch.
2Maintaining deques for max/min values helps in quickly determining if the current window is valid.

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