974. Subarray Sums Divisible by K

Problem #974 (Subarray Sums Divisible by K | Array, Hash Table, Prefix Sum)

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

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Solutions

Approach 1: Hash Table and Prefix Sum

Code

Java

class Solution {
    public int subarraysDivByK(int[] arr, int k) {
        int n = arr.length;
        HashMap<Integer,Integer> map = new HashMap<>();
        int sum = 0;
        int count = 0;
        for(int i = 0; i<n; i++){
            sum+=arr[i];
            if(sum%k==0)
                count++;
            int rem = sum%k;
            if(rem<0)
                rem = rem+k;
            if(map.containsKey(rem))
                count+=map.get(rem);
            map.put(rem,map.getOrDefault(rem,0)+1);
        }
        return count;
    }
}

C++

class Solution {
public:
    int subarraysDivByK(vector<int>& nums, int k) {
        int n = nums.size();
        unordered_map<int, int> map;
        int sum = 0;
        int count = 0;
        for(int i = 0; i < n; i++) {
            sum += nums[i];
            if(sum % k == 0) count++;
            int rem = sum % k;
            if(rem < 0) rem += k;
            if(map[rem]) count += map[rem];
            map[rem] = map[rem] + 1;
        }
        return count;
    }
};

Python3

class Solution:
    def subarraysDivByK(self, nums: List[int], k: int) -> int:
        n = len(nums)
        map = {}
        sum = 0
        count = 0
        for i in range(n):
            sum = sum + nums[i]
            if sum % k == 0: count = count + 1
            rem = sum % k
            if rem < 0: rem = rem + k
            if map.get(rem):
                count = count + map[rem]
                map[rem] = map[rem] + 1
            else:
                map[rem] = 1
        return count

Complexity

• Time: O(n)
• Space: O(n)