Special Numbers

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

Problem Highlights

• 💡 Difficulty: Hard
• ⏰ Time to complete: 25 mins
• 🛠️ Topics: Binary Search, Arrays, Mathematical Logic

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 no such ( x ) exists?
  • A: If no such ( x ) exists that fulfills the condition, the function should return -1.

HAPPY CASE
Input: nums = [3, 6, 7, 7, 0]
Output: 3
Explanation: There are exactly three numbers that are greater than or equal to 3.

EDGE CASE
Input: nums = [0, 0, 0, 0]
Output: -1
Explanation: There is no number ( x ) such that there are ( x ) numbers greater than or equal to ( x ).

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 involves identifying a special index using an optimized search strategy:

• Using binary search on the number of possible valid counts to efficiently determine if such an ( x ) exists.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a binary search to find ( x ) such that there are exactly ( x ) numbers greater than or equal to ( x ) by checking possible values for ( x ).

1) Sort the array.
2) Use binary search on the value range [0, n] where n is the length of the array:
   - For each mid point during the binary search, count how many numbers are greater than or equal to \( mid \).
   - If count equals \( mid \), then \( mid \) is the answer.
   - Adjust the binary search bounds based on whether the count is greater or less than \( mid \).
3) Return -1 if no valid \( x \) is found.

⚠️ Common Mistakes

• Not properly handling the relationship between the indices and the elements, especially in arrays with duplicates or all zeros.

4: I-mplement

Implement the code to solve the algorithm.

def is_special(nums):
    nums.sort()
    n = len(nums)
    left, right = 0, n
    while left < right:
        mid = (left + right) // 2
        # Count how many elements are >= mid
        count = sum(1 for x in nums if x >= mid)
        if count == mid:
            return mid
        elif count > mid:
            left = mid + 1
        else:
            right = mid
    return -1

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 the input [3, 6, 7, 7, 0] to ensure it returns 3.
• Check the edge case [0, 0, 0, 0] to confirm that it correctly returns -1.

6: E-valuate

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

• Time Complexity: O(n log n) due to the sorting step followed by a binary search operation.
• Space Complexity: O(1) since the sorting can be in-place and no additional significant space is used.