Introduction

In this blog post, we will explore how to solve the LeetCode problem "962" 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

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i. Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0. Example 1: Input: nums = [6,0,8,2,1,5] Output: 4 Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): nums[1] = 0 and nums[5] = 5. Example 2: Input: nums = [9,8,1,0,1,9,4,0,4,1] Output: 7 Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): nums[2] = 1 and nums[9] = 1. Constraints: 2 <= nums.length <= 5 104 0 <= nums[i] <= 5 104

Explanation

Here's an efficient solution to find the maximum width ramp in an integer array:

• Preprocessing for Minimums: Create an array min_left where min_left[i] stores the minimum value from nums[0] to nums[i]. This allows us to quickly check if nums[i] is a potential starting point for a ramp.

• Two-Pointer Approach: Use two pointers, left and right. Start left at 0 and right at the end of the array. Move right towards the left, looking for a value nums[right] that is greater than or equal to min_left[left]. If such a right is found, update the maximum width and increment left. If nums[right] is smaller than min_left[left], move right one step to the left until right becomes left.

• Maximize Width: The algorithm iterates, adjusting left and right to find the largest possible difference right - left that satisfies the ramp condition.

• Time and Space Complexity:

  • Runtime: O(n)
  • Storage: O(n)

Code

    def max_width_ramp(nums):
    n = len(nums)
    min_left = [0] * n
    min_left[0] = nums[0]
    for i in range(1, n):
        min_left[i] = min(min_left[i - 1], nums[i])

    max_width = 0
    left = 0
    right = n - 1

    while left < n and right > left:
        if min_left[left] <= nums[right]:
            max_width = max(max_width, right - left)
            left += 1
        else:
            right -= 1

    return max_width