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56 lines
1.6 KiB
Plaintext
56 lines
1.6 KiB
Plaintext
---
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title: '974. Subarray Sums Divisible by K'
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description: Given an integer array nums and an integer k, return the number of non-empty subarrays that have a sum divisible by k
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Hash Table", slug: "hash-table" }
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- { name: "Prefix Sum", slug: "prefix-sum" }
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issue: "https://github.com/prdlk/leetcode/issues/40"
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companies:
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- { name: "Meta", period: "Thirty Days", frequency: 53.8 }
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- { name: "Google", period: "Thirty Days", frequency: 29 }
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- { name: "TikTok", period: "Three Months", frequency: 70.9 }
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- { name: "Amazon", period: "Three Months", frequency: 21.2 }
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- { name: "Microsoft", period: "Six Months", frequency: 24.1 }
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sidebar:
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label: 'Subarray Sums Divisible by K'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `nums = [4,5,0,-2,-3,1], k = 5`
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- Output: `7`
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- 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]
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### Example 2:
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- Input: `nums = [5], k = 9`
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- Output: `0`
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### Constraints:
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- `1 <= nums.length <= 3 * 10^4`
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- `-10^4 <= nums[i] <= 10^4`
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- `2 <= k <= 10^4`
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## Solution
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```py
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class Solution:
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def subarraysDivByK(self, nums: list[int], k: int) -> int:
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remainder_counts = {0: 1}
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cur_sum = 0
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count = 0
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for n in nums:
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cur_sum += n
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remainder = cur_sum % k
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if remainder in remainder_counts:
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count += remainder_counts[remainder]
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remainder_counts[remainder] = remainder_counts.get(remainder, 0) + 1
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return count
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```
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