This approach utilizes the fact that both arrays are sorted. Using two pointers, one for each array, we traverse the arrays to find the smallest common value:

1. Initialize two pointers at the start of each array.
2. Compare the elements pointed by the pointers.
3. If both elements are equal, you've found the minimum common integer.
4. If the element at the first pointer is less than the element at the second pointer, increment the first pointer; otherwise, increment the second pointer.
5. Continue this process until a match is found or one of the pointers reaches the end of its respective array.
6. If no common element is found, return -1.

This approach uses a HashSet to store the elements of the smaller array, providing a quick way to check for common elements:

1. Insert all elements of the smaller array into a HashSet to minimize space usage.
2. Iterate through the larger array and for each element, check if it exists in the HashSet.
3. If a common element is found, return it as the minimum common value due to sorted order.
4. If not, return -1 after iterating through the array.
5. This ensures we keep the lookup time for elements O(1) by using HashSet.