It's basically the remove-k-digits problem but inverted.
Use a greedy approach with a monotonic stack to remove k digits that are smaller than the next digit, ensuring the remaining digits form the largest number. Iterate through the digits, maintaining a stack of digits in decreasing order, and pop when the current digit is larger and removals remain. After processing, if removals remain, remove from the end; then handle 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 maximize the remaining number. Identify edge cases like k=0, k=length, all zeros, and leading zeros after removal.
Select a monotonic stack (or string as stack) to efficiently track digits in decreasing order. Explain why greedy works: at each step, removing a smaller digit before a larger one increases the overall value.
Iterate through each digit: while k>0 and stack top < current digit, pop and decrement k. Push current digit. After iteration, if k>0, remove from the end.
Remove leading zeros from the result. If the result is empty, return '0'. Ensure exactly k digits were removed.
State time complexity O(n) and space O(n). Walk through examples like num='1432219', k=3 to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.