Unit 7 Session 1 (Click for link to problem statements)
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?
-1
indicating that the target is not present.HAPPY CASE
Input: lst = [1, 2, 3, 4, 5], target = 3
Output: 2
Explanation: The target value 3 is located at index 2.
EDGE CASE
Input: lst = [1, 2, 3, 4, 5], target = 6
Output: -1
Explanation: The target value 6 is not in the list.
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 is a classic binary search problem, requiring an iterative approach to searching in a sorted list:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Implement an iterative binary search that finds the index of a target value within a sorted array.
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 the index.
- 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 -1.
⚠️ Common Mistakes
-1
.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 mid
elif lst[mid] > target:
right = mid - 1
else:
left = mid + 1
return -1
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(log n)
because each iteration approximately halves the number of elements to be searched.O(1)
because the iterative approach does not use additional space proportional to the input size.