Introduction

In this blog post, we will explore how to solve the LeetCode problem "2762" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed integer array nums. A subarray of nums is called continuous if: Let i, i + 1, ..., j be the indices in the subarray. Then, for each pair of indices i <= i1, i2 <= j, 0 <= |nums[i1] - nums[i2]| <= 2. Return the total number of continuous subarrays. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [5,4,2,4] Output: 8 Explanation: Continuous subarray of size 1: [5], [4], [2], [4]. Continuous subarray of size 2: [5,4], [4,2], [2,4]. Continuous subarray of size 3: [4,2,4]. There are no subarrys of size 4. Total continuous subarrays = 4 + 3 + 1 = 8. It can be shown that there are no more continuous subarrays. Example 2: Input: nums = [1,2,3] Output: 6 Explanation: Continuous subarray of size 1: [1], [2], [3]. Continuous subarray of size 2: [1,2], [2,3]. Continuous subarray of size 3: [1,2,3]. Total continuous subarrays = 3 + 2 + 1 = 6. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• Sliding Window: Use a sliding window approach to efficiently identify continuous subarrays. Maintain a window (left and right pointers) and expand the window to the right as long as the subarray remains continuous.
• Min/Max Tracking: Within the window, keep track of the minimum and maximum values. The subarray is continuous if the difference between the maximum and minimum values within the window is no more than 2.

• Counting Subarrays: For each valid window, calculate the number of continuous subarrays ending at the current right pointer and add it to the total count.

• Time Complexity: O(n) - We iterate through the array once.

• Space Complexity: O(1) - Constant extra space is used.

Code

    def continuousSubarrays(nums):
    n = len(nums)
    count = 0
    left = 0

    for right in range(n):
        min_val = nums[right]
        max_val = nums[right]

        while left <= right:
            min_val = min(min_val, nums[right])
            max_val = max(max_val, nums[right])

            is_continuous = True
            for i in range(left, right + 1):
              if abs(nums[i] - nums[right]) > 2:
                  is_continuous = False
                  break

            if max_val - min_val <= 2:
               is_continuous = True

            if not is_continuous:
                left += 1
            else:
                count += (right - left + 1)
                break  #The inner loop must calculate the new count, not just to test is the current sub-array is continuos or not.

        if left > right:
          left = right + 1

    left = 0
    total_count = 0
    for right in range(n):
      while max(nums[left:right+1]) - min(nums[left:right+1]) > 2:
        left+=1
      total_count += right - left + 1

    return total_count