<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the number of non-empty <strong>subarrays</strong> that have a sum divisible by </em><code>k</code>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,5,0,-2,-3,1], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 7 subarrays with a sum divisible by k = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5], k = 9
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>2 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def subarraysDivByK(self, nums: List[int], k: int) -> int:
        mp = collections.defaultdict(int)
        psum = 0
        ans = 0

        for num in nums:
            psum += num
            mod = psum % k
       
            look = k - mod
            mp[mod] += 1


        for key in mp:
            if key == 0:
                amount  = mp[key]
                check = (amount+1) * amount/2
                ans += check
            else:
                print(f"key is {key} amount is {mp[key]}")
                amount  = mp[key]
                check = (amount-1) * amount/2
                print(check)
                ans += check
        return int(ans)
21300