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, 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

    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