# Solving Leetcode Interviews in Seconds with AI: Continuous Subarray Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "523" 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 an integer array nums and an integer k, return true if nums has a good subarray or false otherwise. A good subarray is a subarray where:  its length is at least two, and the sum of the elements of the subarray is a multiple of k.  Note that:  A subarray is a contiguous part of the array. An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.    Example 1:  Input: nums = [23,2,4,6,7], k = 6 Output: true Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.  Example 2:  Input: nums = [23,2,6,4,7], k = 6 Output: true Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42. 42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.  Example 3:  Input: nums = [23,2,6,4,7], k = 13 Output: false    Constraints:  1 <= nums.length <= 105 0 <= nums[i] <= 109 0 <= sum(nums[i]) <= 231 - 1 1 <= k <= 231 - 1  

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

*   **Key Idea:** Use the prefix sum modulo k. If two prefix sums have the same remainder when divided by k, then the subarray between them has a sum that's a multiple of k. We use a hash map to store the first occurrence of each remainder.
*   **Optimization:** Store the *index* of the first occurrence of each remainder in the hash map. If we encounter the same remainder again at a later index, check if the distance between the two indices is at least 2.
*   **Complexity:** *Runtime: O(n), Storage: O(min(n, k))*

	
	# Code
	```python
	def checkSubarraySum(nums, k):
    """
    Checks if there is a continuous subarray of size at least two whose elements sum up to a multiple of k.

    Args:
        nums: A list of integers.
        k: An integer.

    Returns:
        True if there is such a subarray, False otherwise.
    """
    remainder_map = {0: -1}  # Initialize with remainder 0 at index -1 to handle the case where the prefix sum itself is a multiple of k
    prefix_sum = 0

    for i, num in enumerate(nums):
        prefix_sum += num
        remainder = prefix_sum % k

        if remainder in remainder_map:
            if i - remainder_map[remainder] >= 2:
                return True
        else:
            remainder_map[remainder] = i

    return False
	```