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Binary Search III
Unit 7 Session X (Click for link to problem statements)
- 💡 Difficulty: Medium
- ⏰ Time to complete: 15 mins
- 🛠️ Topics: Binary Search, Recursive Algorithms, Arrays
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 the function do if the target is not found?
- A: If the target is not found, the function should return -1 to indicate its absence in the list.
HAPPY CASE
Input: lst = [1, 3, 5, 7, 9], target = 5
Output: True
Explanation: The target value 5 exists in the list, so the function returns True.
EDGE CASE
Input: lst = [1, 3, 5, 7, 9], target = 2
Output: False
Explanation: The target value 2 does not exist in the list, so the function returns False.
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 classic application of binary search, using recursion to reduce the search space efficiently:
- Understanding the fundamental properties of binary search.
- Implementing it in a recursive fashion, which is a common approach in algorithmic problem solving.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Develop a recursive binary search function that checks for the presence of a target value within a sorted list.
1) Base Case: If the range is exhausted without finding the target, return False.
2) Calculate the middle index of the current search range.
3) If the middle element is the target, return True.
4) Recurse into the appropriate half of the list based on the comparison between the middle element and the target.
- Not correctly adjusting the search boundaries, which might skip the target or cause unnecessary recursion.
Implement the code to solve the algorithm.
def binary_search_recursive(lst, left, right, target):
if left > right:
return False # Base case: Target not found within bounds
mid = (left + right) // 2
if lst[mid] == target:
return True
elif lst[mid] > target:
return binary_search_recursive(lst, left, mid - 1, target)
else:
return binary_search_recursive(lst, mid + 1, right, target)
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, 3, 5, 7, 9] and a target of 5 to ensure it returns True.
- Validate with a target not in the list (e.g., 2) to check that it returns False.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
-
Time Complexity:
O(log n)
because each recursive call reduces the problem size by half. -
Space Complexity:
O(log n)
due to the recursion depth being equal to the logarithm of the list size.