Binary Search II

Edit PagePage History

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

Problem Highlights

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

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: What should be the return value if the target is not found in the list?
  • A: The function should return -1 indicating that the target is not present in the list.

HAPPY CASE
Input: lst = [1, 2, 3, 4, 5], target = 3
Output: True
Explanation: The target value 3 is found in the list.

EDGE CASE
Input: lst = [1, 2, 3, 4, 5], target = 6
Output: False
Explanation: The target value 6 is not found 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 standard application of the binary search algorithm:

• Applying an iterative binary search to determine if a target exists in a sorted array.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Implement an iterative binary search to check the presence of a target value within a sorted list, returning a boolean result.

1) Initialize pointers for the left and right bounds of the list.
2) While the left pointer is not greater than the right:
   - Calculate the middle index.
   - If the middle element is the target, return True.
   - If the target is less than the middle element, adjust the right pointer to narrow the search to the left half.
   - If the target is greater than the middle element, adjust the left pointer to narrow the search to the right half.
3) If the loop exits without finding the target, return False.

⚠️ Common Mistakes

• Not handling the condition where the target is not found, which should return False.

4: I-mplement

Implement the code to solve the algorithm.

def binary_search(lst, target):
    left, right = 0, len(lst) - 1
    
    while left <= right:
        mid = (left + right) // 2
        if lst[mid] == target:
            return True
        elif lst[mid] > target:
            right = mid - 1
        else:
            left = mid + 1
    
    return False

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, 3, 4, 5] and a target of 3 to ensure it returns True.
• Validate with a target not in the list (e.g., 6) to check that it returns False.

6: E-valuate

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

• Time Complexity: O(log n) because each iteration approximately halves the number of elements to be searched.
• Space Complexity: O(1) because the iterative approach does not use additional space proportional to the input size.