# 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.
``````

``````class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
for (int i = 0; i < nums.size(); ++i) {
int sum = nums[i];
for (int j = i + 1; j < nums.size(); ++j) {
sum += nums[j];
if (sum == k) return true;
if (k != 0 && sum % k == 0) return true;
}
}
return false;
}
};
``````

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

``````class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int n = nums.size(), sum = 0;
unordered_map<int, int> m{{0,-1}};
for (int i = 0; i < n; ++i) {
sum += nums[i];
int t = (k == 0) ? sum : (sum % k);
if (m.count(t)) {
if (i - m[t] > 1) return true;
} else m[t] = i;
}
return false;
}
};
``````

https://discuss.leetcode.com/topic/80975/java-solution

https://discuss.leetcode.com/topic/80793/java-o-n-time-o-k-space/2