## 523. Continuous Subarray Sum

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 the 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.
``````

Note:

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

Thoughts:

1. Use map:
1. Query whether there is a index with value equal to modulo preSum and whether its distance from current i is > 1
2. Use end<modulo preSum: index> to record the modulo preSum value
2. Use only set + add delay
``````class Solution(object):
def checkSubarraySum(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: bool
"""
preS, end = 0, {0: -1}

for i, num in enumerate(nums):
preS += num
if k != 0:
preS %= k
if preS in end: # keep the least index
if i - end[preS] > 1:
return True
else:
end[preS] = i

return False
``````
``````class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int n = nums.size(), sum = 0, pre = 0;
unordered_set<int> modk;
for(int i = 0; i < n; i++){
sum += nums[i];
int mod =  k == 0 ? sum : sum % k;
if(modk.count(mod)) return true;
modk.insert(pre);
pre = mod; // delay for two loops
}
return false;
}

};
``````

Python

``````class Solution(object):
def checkSubarraySum(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: bool
"""
n, s, pre = len(nums), 0, 0
end = set()

for i, num in enumerate(nums):
s += num
if k != 0:
s %= k

if s in end: return True