Harmonizing Two Musical Tracks

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Unit 7 Session 2 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Divide and Conquer, Merge

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• What should be done if one of the tracks is empty?
  • If one track is empty, return the other track as the merged result.
• Are the pitch values distinct?
  • The problem does not specify, so assume the pitch values can have duplicates.
• How should the merged list be ordered?
  • The merged list should be sorted in ascending order.

HAPPY CASE
Input: track1 = [1, 3, 5], track2 = [2, 4, 6]
Output: [1, 2, 3, 4, 5, 6]
Explanation: The tracks are merged in sorted order.

Input: track1 = [10, 20], track2 = [15, 30]
Output: [10, 15, 20, 30]
Explanation: The tracks are merged in sorted order.

EDGE CASE
Input: track1 = [], track2 = [1, 2, 3]
Output: [1, 2, 3]
Explanation: Track1 is empty, so the result is track2.

Input: track1 = [1, 2, 3], track2 = []
Output: [1, 2, 3]
Explanation: Track2 is empty, so the result is track1.

2: M-atch

Match what this problem looks like to known categories of problems, e.g., Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Divide and Conquer problems, we want to consider the following approaches:

• Merge: The problem can be solved by merging two sorted lists into one sorted list. This is a classic merge operation similar to the merge step in merge sort.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a two-pointer approach to merge the two sorted tracks into a single sorted list.

1) Initialize an empty list `merged_track` to store the merged result.
2) Use two pointers, `i` and `j`, initialized to 0, to traverse through `track1` and `track2` respectively.
3) While both pointers are within the bounds of their respective tracks:
    a) Compare the elements at `track1[i]` and `track2[j]`.
    b) Append the smaller element to `merged_track` and increment the corresponding pointer.
4) If there are remaining elements in `track1`, append them to `merged_track`.
5) If there are remaining elements in `track2`, append them to `merged_track`.
6) Return the `merged_track` list as the result.

⚠️ Common Mistakes

• Forgetting to append remaining elements from one of the tracks after one pointer has reached the end.
• Mishandling cases where one or both input lists are empty.

4: I-mplement

Implement the code to solve the algorithm.

def merged_tracks(track1, track2):
    merged_track = []
    i, j = 0, 0
    
    # Merge the two tracks by comparing pitch values
    while i < len(track1) and j < len(track2):
        if track1[i] < track2[j]:
            merged_track.append(track1[i])
            i += 1
        else:
            merged_track.append(track2[j])
            j += 1
    
    # If there are remaining elements in track1, add them
    while i < len(track1):
        merged_track.append(track1[i])
        i += 1
    
    # If there are remaining elements in track2, add them
    while j < len(track2):
        merged_track.append(track2[j])
        j += 1
    
    return merged_track

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with the input [1, 3, 5] and [2, 4, 6]:
  • The merged list should correctly be [1, 2, 3, 4, 5, 6].

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume M and N represent the lengths of track1 and track2, respectively.

• Time Complexity: O(M + N) because each element from both tracks is visited exactly once.
• Space Complexity: O(M + N) because a new list is created to store the merged result.