<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 {
public:
    int subarraysDivByK(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> prefix(n + 1, 0), postfix(n + 1, 0);
        vector<int> bins(k, 0);
        bins[0]++;
        for (int i = 1; i <= n; i++) {
            prefix[i] = prefix[i-1] + nums[i-1];
            int rem = prefix[i] % k;
            if (rem < 0) rem+=k;
            bins[rem]++;
        }
        int res = 0;
        for (int i = n; i >= 0; i--) {
            int targ = prefix[i] % k;
            if (targ < 0) targ += k;
            bins[targ]--;
            res += bins[targ];
            
        }
        return res;
        


    }
};
35000