feat(work): add solution for LeetCode 974 Subarray Sums Divisible by K

work/1/Medium/Array/974.subarray-sums-divisible-by-k.py

@@ -0,0 +1,55 @@
+"""
+974. Subarray Sums Divisible by K
+Difficulty: Medium
+https://leetcode.com/problems/subarray-sums-divisible-by-k/
+
+──────────────────────────────────────────────────
+
+Given an integer array nums and an integer k, return the number of
+non-empty subarrays that have a sum divisible by k.
+
+A subarray is a contiguous part of an array.
+
+
+
+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
+"""
+
+
+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