Continuous Subarray Sum

Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

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

SOLUTION:

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        ctr = 0
        total = [ctr]
        for num in nums:
            ctr += num
            total.append(ctr)
        exists = {}
        for i, f in enumerate(total):
            rem = f % k
            if rem in exists:
                if i - exists[rem] >= 2:
                    return True
            else:
                exists[rem] = i
        return False