feat: refactored docs structure

apps/docs/content/(01-linear-structures)/prefix-sum/(problems)/974_subarray-sums-divisible-by-k.mdx

@@ -0,0 +1,49 @@
+---
+title: '974. Subarray Sums Divisible by K'
+description: Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Prefix Sum", slug: "prefix-sum" }
+issue: "https://github.com/prdlk/leetcode/issues/40"
+sidebar:
+  label: 'Subarray Sums Divisible by K'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [4,5,0,-2,-3,1], k = 5`
+- Output: `7`
+- Explanation: 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]
+
+### Example 2:
+- Input: `nums = [5], k = 9`
+- Output: `0`
+
+### Constraints:
+
+- `1 <= nums.length <= 3 * 10^4`
+- `-10^4 <= nums[i] <= 10^4`
+- `2 <= k <= 10^4`
+
+## Solution
+
+```py
+class Solution:
+    def subarraysDivByK(self, nums: list[int], k: int) -> int:
+        remainder_counts = {0: 1}
+        cur_sum = 0
+        count = 0
+
+        for n in nums:
+            cur_sum += n
+            remainder = cur_sum % k
+
+            if remainder in remainder_counts:
+                count += remainder_counts[remainder]
+
+            remainder_counts[remainder] = remainder_counts.get(remainder, 0) + 1
+
+        return count
+```