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
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