How do you solve Continuous Subarrays on LeetCode?

Continuous Subarrays (LeetCode #2762) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be efficiently solved using the sliding window technique, which allows us to dynamically adjust the range of valid subarrays.

What is the brute force solution for Continuous Subarrays?

The brute force approach for Continuous Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray to see if it meets the condition of having all elements differing by at most 2. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Continuous Subarrays?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases and how your solution handles them. Practice explaining your thought process aloud as you code to simulate the interview environment.

What are common mistakes when solving Continuous Subarrays?

Failing to correctly maintain the left pointer when the condition is violated, leading to incorrect counts. Not considering edge cases where the array length is minimal or where all elements are the same.

#2762

Continuous Subarrays

Medium

ArrayQueueSliding WindowHeap (Priority Queue)Ordered SetMonotonic QueueSliding WindowTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Using the sliding window technique allows us to efficiently find continuous subarrays by maintaining a range of valid elements. This reduces the need for nested loops and checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers (left and right) and a count variable.
2Step 2: Expand the right pointer to include elements in the window while maintaining the condition that the difference between max and min is <= 2.
3Step 3: If the condition fails, move the left pointer to shrink the window until the condition is satisfied again.
4Step 4: For each position of the right pointer, the number of valid subarrays ending at right is (right - left + 1).

solution.py9 lines

1def countContinuousSubarrays(nums):
2    left = 0
3    count = 0
4    n = len(nums)
5    for right in range(n):
6        while max(nums[left:right+1]) - min(nums[left:right+1]) > 2:
7            left += 1
8        count += (right - left + 1)
9    return count

ℹ

Complexity note: The optimal solution runs in O(n) because each element is processed at most twice (once when added and once when removed), making it linear with respect to the input size.

1The problem can be efficiently solved using the sliding window technique, which allows us to dynamically adjust the range of valid subarrays.
2Understanding the relationship between the maximum and minimum values in a subarray is crucial to solving this problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.