974. Subarray Sums Divisible by K LeetCode Solution

In this guide, you will get 974. Subarray Sums Divisible by K LeetCode Solution with the best time and space complexity. The solution to Subarray Sums Divisible by K problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Subarray Sums Divisible by K solution in C++
Subarray Sums Divisible by K solution in Java
Subarray Sums Divisible by K solution in Python
Additional Resources

974. Subarray Sums Divisible by K LeetCode Solution image

Problem Statement of Subarray Sums Divisible by K

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

Complexity Analysis

Time Complexity:
Space Complexity:

974. Subarray Sums Divisible by K LeetCode Solution in C++

class Solution {
 public:
  int subarraysDivByK(vector<int>& nums, int k) {
    int ans = 0;
    int prefix = 0;
    vector<int> count(k);
    count[0] = 1;

    for (const int num : nums) {
      prefix = (prefix + num % k + k) % k;
      ans += count[prefix];
      ++count[prefix];
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

974. Subarray Sums Divisible by K LeetCode Solution in Java

class Solution {
  public int subarraysDivByK(int[] nums, int k) {
    int ans = 0;
    int prefix = 0;
    int[] count = new int[k];
    count[0] = 1;

    for (final int num : nums) {
      prefix = (prefix + num % k + k) % k;
      ans += count[prefix];
      ++count[prefix];
    }

    return ans;
  }
}
// code provided by PROGIEZ

974. Subarray Sums Divisible by K LeetCode Solution in Python

class Solution:
  def subarraysDivByK(self, nums: list[int], k: int) -> int:
    ans = 0
    prefix = 0
    count = [0] * k
    count[0] = 1

    for num in nums:
      prefix = (prefix + num % k + k) % k
      ans += count[prefix]
      count[prefix] += 1

    return ans
 # code by PROGIEZ