README_EN.md

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

Python3

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

Java

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

C++

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

Go

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

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