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# 7. 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
```

## Solution: (Prefix Sum of Remainder)

```cpp
class Solution
{
public:
    bool checkSubarraySum(vector<int> &nums, int k)
    {

        unordered_map<int, int> m;
        int ps = 0;
        m.insert({0, -1});

        for (int i = 0; i < nums.size(); i++)
        {
            ps += nums[i];
            int rem = ps % k;

            if (m.find(rem) != m.end())
            {
                int dif = i - m[rem];
                if (dif >= 2)
                {
                    return true;
                }
            }
            m.insert({rem, i});
        }

        return false;
    }
};
```

**Time Complexity: O(n)**