Another method to tackle this issue involves the use of pointers. If the last element of the first array is greater than that of the second array, I will exchange them. Next, I will decrement the pointer in the first array by one and make a comparison again, continuing this process until every element in the first array is smaller than those in the second. Finally, I will sort both arrays. The time complexity for this method is comparable to the previous solution, at O(M*N log M), since sorting is necessary after each swap. However, this method may yield better performance in practice if the two initial arrays are nearly sorted, as it may require fewer swaps.