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?
HAPPY CASE
Input: [0, 0, 0, 0, 1, 1, 1]
Output: 4
Explanation: There are four 0's before the first 1 appears.
EDGE CASE
Input: [1, 1, 1, 1, 1]
Output: 0
Explanation: There are no 0's in the array, so the function returns 0.
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 an efficient counting challenge best suited for a binary search technique:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Utilize an iterative binary search to find the index of the first occurrence of 1 in the sorted array.
1) Initialize low to 0 and high to the length of the array minus one.
2) While low is less than or equal to high:
- Calculate the mid index.
- Check if the mid index is the first occurrence of 1:
- If nums[mid] is 1 and nums[mid - 1] is 0 (or mid is 0), then count of zeros is mid.
- If nums[mid] is 0, move the low to mid + 1.
- Otherwise, move the high to mid - 1.
3) If 1 is never found, return the array length
⚠️ Common Mistakes
Implement the code to solve the algorithm.
def count_zeros(nums):
low, high = 0, len(nums) - 1
while low <= high:
mid = (low + high) // 2
if lst[mid] == 1:
if mid == 0 or lst[mid - 1] == 0:
return mid # Number of 0's is the index of the first '1'
high = mid - 1
else:
low = mid + 1
return len(nums) # If no '1' is found, all are zeros
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)
due to the binary search process.O(1)
since no additional space is used beyond variables for the search indices.