## DEV Community

Abhishek Chaudhary

Posted on

# 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
``````