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Find Missing
Unit 7 Session 1 (Click for link to problem statements)
- 💡 Difficulty: Medium
- ⏰ Time to complete: 15 mins
- 🛠️ Topics: Binary Search, 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 happens if there are no missing numbers?
- A: If no numbers are missing, the missing number would logically be ( n ), since the numbers are from 0 to ( n-1 ).
HAPPY CASE
Input: nums = [0, 1, 3, 4, 5]
Output: 2
Explanation: The number 2 is missing in the sequence.
EDGE CASE
Input: nums = [0, 1, 2, 3, 4]
Output: 5
Explanation: Since no numbers are missing from 0 to 4, the missing number is 5.
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 variant of the classic binary search:
- Adapting binary search to detect an inconsistency in an otherwise complete series of sequential numbers.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Use binary search to efficiently find the missing number in a series that is nearly complete.
1) Establish binary search bounds with `left` at 0 and `right` at the last index of the array.
2) While `left` is less than `right`:
- Compute the middle index.
- If the element at the middle index matches the index, adjust `left` to `mid + 1`.
- Otherwise, adjust `right` to `mid`.
3) The missing number will be at the position indicated by `left` after the loop completes.
- Not considering the possibility that the missing number could be at the very end of the sequence (i.e., ( n )).
Implement the code to solve the algorithm.
def find_missing(nums):
left, right = 0, len(nums)
while left < right:
mid = (left + right) // 2
if nums[mid] > mid:
right = mid
else:
left = mid + 1
return left
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Test with input [0, 1, 3, 4, 5] to confirm it finds the missing number 2.
- Check with input [0, 1, 2, 3, 4] to ensure it returns 5, demonstrating that it correctly identifies the next logical missing number.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
-
Time Complexity:
O(log n)
because each iteration of the loop narrows the search space by half. -
Space Complexity:
O(1)
as the space used does not scale with the size of the input.