Question

Suppose we are given two \(n\) -element sorted sequences \(A\) and \(B\) that should not be viewed as sets (that is, \(A\) and \(B\) may contain duplicate entries). Describe an \(O(n)\) -time method for computing a sequence representing the \(\operatorname{set} A \cup B\) (with no duplicates).

Short answer

Use two pointers to merge sequences and append unique elements in O(n) time.

Step-by-step solution

- Initialize pointers and result set

Create two index pointers, one for each sequence, `i` for sequence `A` and `j` for sequence `B`. Also, create an empty list `C` to store the final result without duplicates.

- Iterate through sequences

Start a while loop that continues until both pointers `i` and `j` have scanned all elements in their respective sequences. This means the loop will run while `i < n` or `j < n`.

- Compare and append unique elements

Inside the loop, compare the elements at indices `i` and `j` in sequences `A` and `B`, respectively. If `A[i] < B[j]`, append `A[i]` to `C` if it's not already the last element in `C`, then increment `i`. If `A[i] > B[j]`, append `B[j]` to `C` if it's not already the last element in `C`, then increment `j`. If `A[i] == B[j]`, append either `A[i]` or `B[j]` to `C` (but only once) and increment both `i` and `j`.

- Append remaining elements

After the loop, handle any remaining elements in either sequence (as one of them might be fully iterated over earlier). Append these remaining elements to `C` if they are not already the last element in `C`.

- Result

The list `C` now contains all unique elements from sequences `A` and `B` in sorted order. This is the desired result.

Key concepts

time complexity

Understanding time complexity is vital when analyzing how efficient an algorithm is. In our exercise, we are given two sorted sequences, each with n elements. The goal is to merge these sequences into a new sequence with unique elements.

The time complexity of our method is O(n), which means the algorithm runs in linear time proportional to the size of the sequences.
Every element from both sequences is processed once, making each operation constant time.

This efficiency comes from the fact that we make one pass through each sequence, comparing and appending elements to the result list. Unlike more complex algorithms such as quadratic time algorithms (O(n^2)), linear time algorithms (O(n)) are optimal for large datasets.

In summary, by iterating through both sequences simultaneously while maintaining the order, we ensure that our solution is efficient and scalable for large n.

algorithm design

Algorithm design is all about finding the most effective way to solve a problem. In this exercise, the approach leverages the sorted nature of the input sequences.

Here's a breakdown of the algorithm:

• Initialize index pointers for both sequences and an empty result list
• Use a while loop to iterate through both sequences
• Compare elements and append unique values to the result list
• Handle any remaining elements in either sequence

The core idea is to use two pointers, one for each sequence. We compare the elements at these pointers and append the smaller element to the result list. If the elements are equal, we add it once and move both pointers forward. This process naturally merges the sequences while removing duplicates.

Handling remaining elements ensures that if one sequence is fully processed before the other, the algorithm still captures all unique elements.
Essentially, by combining these strategies, we ensure minimal operations, leading to a clear and efficient algorithm.

duplicate removal

Duplicate removal is crucial when combining sets or sequences. In our exercise, removing duplicates while preserving the order of elements is key. Here’s how the algorithm achieves this:

• Compare elements at each step, only appending them if they are not already in the result list
• When encountering equal elements in both sequences, append it once and move both pointers

To be specific, inside the loop, if `A[i]` is less than `B[j]`, we check if `A[i]` is already the last element in the result list. If not, we append it. If `A[i]` equals `B[j]`, we add it and move both pointers. This ensures only unique elements make it to the final result.

The final step deals with remaining elements in either sequence. By checking if they are not already in the result list, duplicate entries are avoided even in these leftover elements.

Removing duplicates efficiently ensures the final merged sequence maintains only unique values, fulfilling the requirement of representing a set.