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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.

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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

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