Question

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.

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 * 104
• -104 <= nums[i] <= 104
• 2 <= k <= 104

Solutions

Solution 1

class Solution:
  def subarraysDivByK(self, nums: List[int], k: int) -> int:
    cnt = Counter({0: 1})
    ans = s = 0
    for x in nums:
      s = (s + x) % k
      ans += cnt[s]
      cnt[s] += 1
    return ans

class Solution {
  public int subarraysDivByK(int[] nums, int k) {
    Map<Integer, Integer> cnt = new HashMap<>();
    cnt.put(0, 1);
    int ans = 0, s = 0;
    for (int x : nums) {
      s = ((s + x) % k + k) % k;
      ans += cnt.getOrDefault(s, 0);
      cnt.merge(s, 1, Integer::sum);
    }
    return ans;
  }
}

class Solution {
public:
  int subarraysDivByK(vector<int>& nums, int k) {
    unordered_map<int, int> cnt{{0, 1}};
    int ans = 0, s = 0;
    for (int& x : nums) {
      s = ((s + x) % k + k) % k;
      ans += cnt[s]++;
    }
    return ans;
  }
};

func subarraysDivByK(nums []int, k int) (ans int) {
  cnt := map[int]int{0: 1}
  s := 0
  for _, x := range nums {
    s = ((s+x)%k + k) % k
    ans += cnt[s]
    cnt[s]++
  }
  return
}

function subarraysDivByK(nums: number[], k: number): number {
  const counter = new Map();
  counter.set(0, 1);
  let s = 0,
    ans = 0;
  for (const num of nums) {
    s += num;
    const t = ((s % k) + k) % k;
    ans += counter.get(t) || 0;
    counter.set(t, (counter.get(t) || 0) + 1);
  }
  return ans;
}