The monotonic stack solution clicked for me pretty fast, but I fumbled the edge cases.
Use a greedy approach with a monotonic stack to remove digits that are larger than the next digit, ensuring the smallest possible number. Iterate through the digits, maintaining a stack of selected digits, and pop from the stack when the current digit is smaller and removals are still allowed. After processing, handle any remaining removals by trimming from the end, and finally strip leading zeros.
Pro tip: Clarify edge cases upfront, such as when k equals the length of the string (result should be '0') or when leading zeros appear after removal. Also, discuss time and space complexity (O(n) time, O(n) space) to demonstrate thoroughness.
Restate the problem to ensure clarity: remove exactly k digits to minimize the number while preserving order and avoiding leading zeros. Identify edge cases like k=0, k=length, all digits same, and strings with zeros.
Select a monotonic stack (or deque) to efficiently track digits to keep. Explain that the greedy strategy of removing a digit when it is greater than the next digit leads to the smallest number.
Iterate through each digit, and while the stack is not empty, the top of the stack is greater than the current digit, and removals remain, pop from the stack. Then push the current digit. After iteration, if removals remain, remove from the end of the stack.
Convert the stack to a string, strip leading zeros, and if the result is empty, return '0'. Ensure exactly k digits were removed.
State that the algorithm runs in O(n) time and O(n) space. Walk through a few test cases to verify correctness, including edge cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.