# Solving Leetcode Interviews in Seconds with AI: Maximum Width Ramp


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```