Sure, here is a step-by-step approach to solve the problem:

1. Initialize a variable `count` to 0, which will keep track of the number of subarrays whose sum is divisible by K.

2. Create a dictionary `remainder_count` to store the count of remainders obtained when the cumulative sum of elements in the array is divided by K. Initialize `remainder_count` with a single key-value pair: (0, 1), as the cumulative sum of an empty subarray is 0.

3. Initialize a variable `cumulative_sum` to 0, which will store the cumulative sum of elements in the array.

4. Iterate through each element `num` in the array A.

5. Update the `cumulative_sum` by adding `num` to it.

6. Calculate the remainder `remainder` when `cumulative_sum` is divided by K.

7. If `remainder` is negative, add K to it to make it positive. This is done to handle negative remainders.

8. If `remainder` exists in `remainder_count`, increment `count` by the value associated with `remainder` in `remainder_count`.

9. If `remainder` does not exist in `remainder_count`, add it to `remainder_count` with a value of 1.

10. Increment the value associated with `remainder` in `remainder_count` by 1.

11. After iterating through all the elements in the array, print the value of `count`, which represents the number of subarrays whose sum is divisible by K.

Here is a sample implementation in Python:

```python
def count_subarrays(A, K):
count = 0
remainder_count = {0: 1}
cumulative_sum = 0

for num in A:
cumulative_sum += num
remainder = cumulative_sum % K

if remainder

Question

Sure, here is a step-by-step approach to solve the problem:

1. Initialize a variable `count` to 0, which will keep track of the number of subarrays whose sum is divisible by K.

2. Create a dictionary `remainder_count` to store the count of remainders obtained when the cumulative sum of elements in the array is divided by K. Initialize `remainder_count` with a single key-value pair: (0, 1), as the cumulative sum of an empty subarray is 0.

3. Initialize a variable `cumulative_sum` to 0, which will store the cumulative sum of elements in the array.

4. Iterate through each element `num` in the array A.

5. Update the `cumulative_sum` by adding `num` to it.

6. Calculate the remainder `remainder` when `cumulative_sum` is divided by K.

7. If `remainder` is negative, add K to it to make it positive. This is done to handle negative remainders.

8. If `remainder` exists in `remainder_count`, increment `count` by the value associated with `remainder` in `remainder_count`.

9. If `remainder` does not exist in `remainder_count`, add it to `remainder_count` with a value of 1.

10. Increment the value associated with `remainder` in `remainder_count` by 1.

11. After iterating through all the elements in the array, print the value of `count`, which represents the number of subarrays whose sum is divisible by K.

Here is a sample implementation in Python:

```python
def count_subarrays(A, K):
    count = 0
    remainder_count = {0: 1}
    cumulative_sum = 0

for num in A:
        cumulative_sum += num
        remainder = cumulative_sum % K

if remainder < 0:
            remainder += K

if remainder in remainder_count:
            count += remainder_count[remainder]

remainder_count[remainder] = remainder_count.get(remainder, 0) + 1

return count

# Example usage
A = [4, 5, 0, -2, -3, 1]
K = 5
result = count_subarrays(A, K)
print(result)  # Output: 7
```

This implementation has a time complexity of O(N), where N is the length of the array A.

Knowee AI · Accepted Answer