Introduction

In this blog post, we will explore how to solve the LeetCode problem "795" 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

Given an integer array nums and two integers left and right, return the number of contiguous non-empty subarrays such that the value of the maximum array element in that subarray is in the range [left, right]. The test cases are generated so that the answer will fit in a 32-bit integer. Example 1: Input: nums = [2,1,4,3], left = 2, right = 3 Output: 3 Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3]. Example 2: Input: nums = [2,9,2,5,6], left = 2, right = 8 Output: 7 Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 109 0 <= left <= right <= 109

Explanation

Here's the breakdown of the solution:

• Sliding Window Approach: The core idea is to use a sliding window technique. We maintain two pointers, start and end, and iterate through the array. The goal is to count subarrays ending at end that satisfy the condition.
• Two Cases: We consider two scenarios: 1) nums[end] is within the range [left, right]. In this case, all subarrays ending at end and starting after the last element exceeding right are valid. 2) nums[end] is greater than right. This resets the valid subarray count.

• Efficient Counting: We need to keep track of the index of the last element that exceeded right to calculate the valid subarrays efficiently.

• Runtime Complexity: O(n) & Storage Complexity: O(1)

Code

    def numSubarrayBoundedMax(nums, left, right):
    """
    Counts the number of contiguous non-empty subarrays such that the value of the maximum
    array element in that subarray is in the range [left, right].

    Args:
        nums (List[int]): The input array of integers.
        left (int): The lower bound of the range.
        right (int): The upper bound of the range.

    Returns:
        int: The number of subarrays that meet the requirements.
    """
    count = 0
    valid_subarray_count = 0
    last_invalid = -1

    for i in range(len(nums)):
        if left <= nums[i] <= right:
            valid_subarray_count = i - last_invalid
            count += valid_subarray_count
        elif nums[i] < left:
            count += valid_subarray_count
        else:
            valid_subarray_count = 0
            last_invalid = i

    return count