🚀Easy C Language Solution Using Suitable Array Concepts🚀
IntuitionThe function calculates the number of pairs (nums1[i], nums2[j]) where the element from nums1 is divisible by k times the element from nums2. This invo
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The function calculates the number of pairs (nums1[i], nums2[j]) where the element from nums1 is divisible by k times the element from nums2. This involves checking each pair of elements from both arrays and applying the divisibility condition.
# Approach
<!-- Describe your approach to solving the problem. -->
1. **Initialize Count:**
Start with a count variable set to 0. This will keep track of valid pairs.

2. **Iterate Through Both Arrays:**

3. **Use a nested loop:**
Outer loop iterates over elements of nums1.
Inner loop iterates over elements of nums2.

4. **Check Divisibility:**
For each pair (nums1[i], nums2[j]), compute nums2[j] * k.
Check if nums1[i] % (nums2[j] * k) == 0 (i.e., nums1[i] is divisible by the scaled value).
If the condition is true, increment the count.

5. **Return Result:**
After iterating through all pairs, return the total count.
# Complexity
- Time complexity: O(n1.n2)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```c []
int numberOfPairs(int* nums1, int nums1Size, int* nums2, int nums2Size, int k) {
    int count = 0;
    for(int i = 0; i <= nums1Size - 1; i++)
    {
        for(int j = 0; j <= nums2Size - 1; j++)
        {
            if(nums1[i] % (nums2[j] * k) == 0)
            {
                count++;
            }
        }
    }
    return count;
}
```
['Array', 'Hash Table', 'C']