# Solving Leetcode Interviews in Seconds with AI: Minimum Size Subarray in Infinite Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2875" 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
	> You are given a 0-indexed array nums and an integer target. A 0-indexed array infinite_nums is generated by infinitely appending the elements of nums to itself. Return the length of the shortest subarray of the array infinite_nums with a sum equal to target. If there is no such subarray return -1.   Example 1:  Input: nums = [1,2,3], target = 5 Output: 2 Explanation: In this example infinite_nums = [1,2,3,1,2,3,1,2,...]. The subarray in the range [1,2], has the sum equal to target = 5 and length = 2. It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.  Example 2:  Input: nums = [1,1,1,2,3], target = 4 Output: 2 Explanation: In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...]. The subarray in the range [4,5], has the sum equal to target = 4 and length = 2. It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.  Example 3:  Input: nums = [2,4,6,8], target = 3 Output: -1 Explanation: In this example infinite_nums = [2,4,6,8,2,4,6,8,...]. It can be proven that there is no subarray with sum equal to target = 3.    Constraints:  1 <= nums.length <= 105 1 <= nums[i] <= 105 1 <= target <= 109  

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

*   **High-Level Approach:**
    *   Calculate the sum of the `nums` array. If the target is less than the smallest element in `nums` and the sum of `nums` is greater than the target, we can immediately return `-1`. Otherwise, use a sliding window approach to find the shortest subarray within `nums` that sums to `target`, and calculate how many full `nums` array repetitions are required to reach the target sum.
    *   Consider the case when target is perfectly divisible by the sum of nums.

*   **Complexity:**
    *   Runtime: O(n), where n is the length of the nums array. Storage: O(1).

	
	# Code
	```python
	def shortestSubarray(nums, target):
    n = len(nums)
    total_sum = sum(nums)
    min_len = float('inf')

    if target < min(nums) and total_sum > target:
        return -1

    if total_sum == 0:
        if target == 0:
            return 1
        else:
            return -1

    num_repetitions = max(0, target // total_sum)
    remaining_target = target % total_sum

    if remaining_target == 0:
        if total_sum != 0:
             return num_repetitions * n
        else:
            return -1

    combined_nums = nums * 2  # Concatenate nums with itself to handle wraparound
    current_sum = 0
    window_start = 0

    for window_end in range(2 * n):
        current_sum += combined_nums[window_end]

        while current_sum > remaining_target or abs(current_sum - remaining_target) < 1e-9:
           
            if current_sum == remaining_target:
                min_len = min(min_len, window_end - window_start + 1)
                
            current_sum -= combined_nums[window_start]
            window_start += 1
            if current_sum < remaining_target:
                 break
           
        if current_sum == remaining_target:
             min_len = min(min_len, window_end - window_start + 1)


    if min_len == float('inf'):
        return -1
    else:
        return min_len + num_repetitions * n
	```