Where Does it Go (Iterative)

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

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 15 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 happen if the target is greater than all elements in the array?
  • A: The function should return the length of the array, indicating that the target should be inserted at the end.

HAPPY CASE
Input: nums = [1, 3, 5, 6], target = 5
Output: 2
Explanation: 5 already exists in the array at index 2.

EDGE CASE
Input: nums = [1, 3, 5, 6], target = 7
Output: 4
Explanation: 7 does not exist in the array but would fit at the end, so the index 4 is returned.

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 fundamental application of binary search:

• Adjusting binary search to not only find elements but also to find where an element should be if it's not present.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use an iterative binary search approach to find either the position of the target or where it would be inserted if not found.

1) Initialize `left` to 0 and `right` to the length of the array minus one.
2) While `left` is less than or equal to `right`:
   - Calculate the middle index.
   - If the middle element is the target, return the middle index.
   - If the target is less than the middle element, adjust `right` to `mid - 1`.
   - Otherwise, adjust `left` to `mid + 1`.
3) If the target is not found, `left` will be at the index where the target should be inserted.

⚠️ Common Mistakes

• Returning the wrong index when the element is not found.
• Overlooking the conditions that adjust the left and right pointers, leading to incorrect results.

4: I-mplement

Implement the code to solve the algorithm.

def search_insert(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return left

5: R-eview

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

• Test the function with input [1, 3, 5, 6] and target 5 to ensure it finds index 2.
• Validate with target 7 to check that it correctly identifies index 4 as the insertion point.

6: E-valuate

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

• Time Complexity: O(log n) due to the binary search approach.
• Space Complexity: O(1) as it requires a constant amount of space.