Sponsored
Sponsored
This approach leverages prefix sums and the properties of modular arithmetic to efficiently count subarrays whose sums are divisible by k. By maintaining a hashmap (or dictionary) that keeps track of the frequency of each remainder observed when computing prefix sums modulo k, we can determine how many valid subarrays end at each position in the array.
For any position in the array, if we have the same remainder as a previously computed prefix sum, it indicates that the subarray between the two indices is divisible by k.
Time Complexity: O(n), where n is the number of elements in the array, as we iterate over the array once.
Space Complexity: O(k), where k is the number of different remainders we track in our modCount array.
1#include <stdio.h>
2#include <stdlib.h>
3
4int subarraysDivByK(int* nums, int numsSize, int k) {
5
In this C implementation, an array is used to store the frequency of each remainder when the cumulative sum is divided by k. We initialize the first element to 1 as a subarray with sum 0 (no elements) is considered valid when dividing by k. We then iterate over the input array, updating the cumulative sum and calculating the current remainder. Each time we find a previously seen remainder in our modCount array, it indicates subarrays ending at the current position are divisible by k.
The brute force approach involves checking all possible subarrays to find those whose sum is divisible by k. This method is less efficient and should be used primarily for small inputs due to its higher time complexity. We iterate over all subarray starting points and compute the sum for all possible ending positions, counting subarrays that meet the criteria.
Time Complexity: O(n^2), where n is the number of elements in the array (due to double loop).
Space Complexity: O(1), as only integer counters and accumulators are used.
1#include <vector>
int subarraysDivByK(std::vector<int>& nums, int k) {
int count = 0;
int n = nums.size();
for (int start = 0; start < n; start++) {
int sum = 0;
for (int end = start; end < n; end++) {
sum += nums[end];
if (sum % k == 0) {
count++;
}
}
}
return count;
}
int main() {
std::vector<int> nums = {4, 5, 0, -2, -3, 1};
int k = 5;
std::cout << "Number of subarrays divisible by " << k << ": " << subarraysDivByK(nums, k) << std::endl;
return 0;
}Implemented in C++, this approach iteratively evaluates each subarray by computing sums and checking divisibility by k. This brute force method involves a pair of nested loops, and directly sums elements between two indices to check if the total is divisible by k.