Let \(S\) and \(T\) be two arrays of \(m\) and \(n\) elements, respectively. Write an algorithm that finds all the common elements and stores them in an array U. Show that this can be done in \(\Theta(n+m)\) time.

Understanding the time complexity of an algorithm is crucial for evaluating its efficiency. It represents how the execution time of an algorithm increases with the input size. When an algorithm is said to have a time complexity of \(\Theta(n+m)\), it implies that the number of operations needed to complete the algorithm is directly proportional to the sum of the sizes of the two input arrays.

For our algorithm that finds common elements between two arrays, we scan through each array once, comparing elements and possibly storing shared values into a third array. Since each comparison involves only one element from each array, the number of comparison operations will be at most the sum of the lengths of the two arrays, \(m\) and \(n\). This linear relationship is why the time complexity is \(\Theta(n+m)\). It's worth noting that this type of time complexity is highly efficient for large data sets, as it does not depend on elements being compared multiple times, as would be the case with a nested loop, which would result in a time complexity of \(\Theta(n*m)\).

The implementation of an algorithm involves a series of steps that systematically solve the problem at hand. For our common elements finder, the algorithmic steps start with initializing pointers that will be used to traverse the given arrays.

The core process involves looping through the arrays and comparing their elements, as described in our solution. Here, the careful increment of pointers ensures that each array is processed efficiently. If an element in array \(S\) is lower than the corresponding element in array \(T\), we only increment the pointer on \(S\) because we know that no element in \(T\) before its current position could be a match. This stepwise approach capitalizes on the fact that common elements will appear in the same order in both \(S\) and \(T\) if the arrays are sorted, which is a usual precondition for this type of algorithm.

Pseudocode helps visualize the structure of an algorithm in a simplified, language-agnostic manner. It is vital for planning the logic before the actual coding starts. Below is the pseudocode for finding common elements in arrays \(S\) and \(T\), stored into array \(U\):

\begin{align*}\ &\text{Initialize } i, j, k \text{ to } 0 \ &\text{while } (i < m \text{ AND } j < n) \{\ &\quad \text{if } (S[i] == T[j]) \{\ &\quad\quad U[k] \leftarrow S[i] \ &\quad\quad i++, j++, k++ \ &\quad \} \text{ else if } (S[i] < T[j]) \{\ &\quad\quad i++ \ &\quad \} \text{ else } \{\ &\quad\quad j++ \ &\quad \}\ &\}\ &\text{End while}\end{align*}\
The pseudo code is intentionally designed to be reflective of the step-by-step solution, displaying how each line directly corresponds to an action to be performed in order. This allows anyone reading it to understand the flow of operations and the logic behind each decision, preparing them for translating it into actual programming code.