Time Complexity: O(m * n), where m is the size of array A and n is the size of array B. This is because we are iterating through each element of A and for each element, we are iterating through B.

Space Complexity: O(n), where n is the size of array B. This is due to the visited array that we are using to keep track of which elements in B have been used.

Optimal Approach

Algorithm

Initialize two pointers: i = 0 (for array A) and j = 0 (for array B).
If A[i] < B[j], increment i since no further matching is possible for A[i] in array B.
If A[i] == B[j], it's a match. Add A[i] to the result list and increment both i and j.
If A[i] > B[j], increment j since no further matching is possible for B[j] in array A.
Continue until you traverse either array A or B completely.

Code

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