# 523. Continuous Subarray Sum

#### Medium

***

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:**

<pre><code>Input: nums = [23,2,4,6,7], k = 6
<strong>Output:
</strong> true
<strong>Explanation:
</strong> [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
</code></pre>

**Example 2:**

<pre><code>Input: nums = [23,2,6,4,7], k = 6
<strong>Output:
</strong> true
<strong>Explanation:
</strong> [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.
</code></pre>

**Example 3:**

<pre><code>Input: nums = [23,2,6,4,7], k = 13
<strong>Output:
</strong> false
</code></pre>

&#x20;

**Constraints:**

* `1 <= nums.length <= 105`
* `0 <= nums[i] <= 109`
* `0 <= sum(nums[i]) <= 231 - 1`
* `1 <= k <= 231 - 1`

```python
class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        d = {0:0}
        s = 0
        for index, num in enumerate(nums):
            s += num
            if s % k not in d:
                d[s%k] = index+1
            elif d[s%k] < index:
                return True
        return False
```