Unit 7 Session 2 (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?
x
exists?
x
exists that fulfills the condition, the function should return -1.HAPPY CASE
Input: nums = [3, 6, 7, 7, 0]
Output: 4
Explanation: There are exactly four 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.
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:
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) 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 \( mid \).
2) Return -1 if no valid x is found.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
def is_special(nums):
n = len(nums)
# Traverse from x = 0 to x = len(nums)
for x in range(n + 1):
# Find the position in nums where the elements start being >= x
# Since the list is sorted, we can use the position directly to find how many elements are >= x
# Binary search can be used here for optimization
low, high = 0, n
while low < high:
mid = (low + high) // 2
if nums[mid] >= x:
high = mid
else:
low = mid + 1
# low is now the first index where nums[low] >= x
# The number of elements >= x is n - low
if n - low == x:
return x
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(n log n)
due to the sorting step followed by a binary search operation.O(1)
since the sorting can be in-place and no additional significant space is used.