# Solving Leetcode Interviews in Seconds with AI: Patching Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "330" 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
	> Given a sorted integer array nums and an integer n, add/patch elements to the array such that any number in the range [1, n] inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.   Example 1:  Input: nums = [1,3], n = 6 Output: 1 Explanation: Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4. Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3]. Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6]. So we only need 1 patch.  Example 2:  Input: nums = [1,5,10], n = 20 Output: 2 Explanation: The two patches can be [2, 4].  Example 3:  Input: nums = [1,2,2], n = 5 Output: 0    Constraints:  1 <= nums.length <= 1000 1 <= nums[i] <= 104 nums is sorted in ascending order. 1 <= n <= 231 - 1  

	# Explanation
	Here's an approach to solve the problem efficiently:

*   **Greedy Approach:** Maintain a `reachable` variable, representing the largest number currently reachable. Iterate through the `nums` array. If the current number in `nums` is less than or equal to `reachable + 1`, we can extend our reachable range. If not, we need to patch a number equal to `reachable + 1` to extend our range.
*   **Patching Strategy:** When patching, always patch with `reachable + 1`. This extends the reachable range to `reachable + (reachable + 1) = 2 * reachable + 1`, which is the maximum range extension possible with a single patch.
*   **Handling Large n:** Keep patching until the `reachable` range covers `n`.

*   **Time & Space Complexity:** O(m + log n) where m is the length of nums, O(1)
Here's the Python code:

	
	# Code
	```python
	def minPatches(nums, n):
    patches = 0
    reachable = 0
    i = 0
    while reachable < n:
        if i < len(nums) and nums[i] <= reachable + 1:
            reachable += nums[i]
            i += 1
        else:
            reachable += reachable + 1
            patches += 1
    return patches
	```