Ternary Search

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Unit 7 Session 1 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20 mins
• 🛠️ Topics: Ternary Search, Arrays, Iterative Algorithms

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?

• Q: How does the algorithm handle when the target value does not exist in the array?
  • A: The function should return -1 if the target is not found.

HAPPY CASE
Input: lst = [1, 2, 5, 8, 12, 15, 18, 22, 27], target = 15
Output: 5
Explanation: The target 15 is located at index 5.

EDGE CASE
Input: lst = [1, 2, 5, 8, 12, 15, 18, 22, 27], target = 20
Output: -1
Explanation: The target value 20 is not in the list.

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.

This problem is a variation of binary search known as ternary search:

• The array is divided into three parts, and the target is searched in the segment where it may exist based on comparisons.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Implement an iterative ternary search function that divides the search space into three parts instead of two, as in binary search.

1) Initialize pointers for the low and high bounds of the array.
2) While the low pointer is less than or equal to the high:
   - Calculate two mid points, mid1 and mid2, that divide the array into three parts.
   - Compare the target with the elements at mid1 and mid2:
     - If the target matches mid1 or mid2, return the index.
     - Adjust pointers based on whether the target is less than mid1, between mid1 and mid2, or greater than mid2.
3) If the loop exits without finding the target, return -1.

⚠️ Common Mistakes

• Incorrectly calculating mid1 and mid2 such that they do not properly divide the array into three parts.
• Not handling the comparisons correctly which could skip over the target.

4: I-mplement

Implement the code to solve the algorithm.

def ternary_search(lst, target):
    low, high = 0, len(lst) - 1
    
    while low <= high:
        # Divide the range into three parts
        mid1 = low + (high - low) // 3
        mid2 = high - (high - low) // 3
        
        # Check if the target is found at any mid
        if lst[mid1] == target:
            return mid1  # Target found at mid1
        if lst[mid2] == target:
            return mid2  # Target found at mid2
        
        # Narrow down the search range based on the comparison
        if target < lst[mid1]:
            high = mid1 - 1  # Target in the first third
        elif target > lst[mid2]:
            low = mid2 + 1  # Target in the third third
        else:
            low = mid1 + 1  # Target in the middle third
            high = mid2 - 1
    
    return -1  # Target not found

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 a list [1, 2, 5, 8, 12, 15, 18, 22, 27] and a target of 15 to ensure it identifies the correct index.
• Validate with a target not in the list (e.g., 20) to check that it returns -1.

6: E-valuate

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

• Time Complexity: O(log_3 n) (base 3), because it divides the search space into thirds each time.
• Space Complexity: O(1), because the solution only uses a few variables to store the indices and comparison results, regardless of the size of the input array.