# Continuous Subarray Sum SOLUTION

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### 523. Continuous Subarray Sum SOLUTION

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to a multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Constraints:

The length of the array won’t exceed 10,000.
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

SOLUTION:
``````class Solution:
def checkSubarraySum(self, nums: List[int], k: int) -> bool:
if len(nums) < 2: # need a minimum len of 2
return False

if k == 1: # always true
return True

d = {}
s = 0

for i,n in enumerate(nums):
s += n
check_val = s % k if k else s
if check_val in d:
if i - d[check_val] > 1: # need minimum length of 2
return True
else:
d[check_val] = i

check_val = s % k if k else s # edge-case when the entire array sum is divisible by k but none of the prefix sums are
return check_val == 0 or False``````