Use a two-pointer technique: one pointer (write) tracks the position for the next unique element, and the other (read) scans the array. Since the array is sorted, duplicates are adjacent, so compare the current element with the last written unique element and copy only when different. Return the write pointer as the new length.
Pro tip: Explicitly state the invariants: the prefix up to write-1 contains all unique elements in order, and read >= write always. This demonstrates rigorous thinking and makes the O(n) time and O(1) space justification trivial.
Restate the problem to ensure understanding: in-place modification, return new length, preserve order, O(n) time and O(1) space. Ask if empty array or single element is possible.
Explain that you'll maintain a write index starting at 1 (since first element is always unique) and iterate read from 1 to n-1. When nums[read] != nums[write-1], copy nums[read] to nums[write] and increment write.
Trace the algorithm on a small example like [1,1,2,3,3] to show how the array is modified and why the prefix remains correct. Highlight that elements beyond the new length are irrelevant.
Justify O(n) time: each element is visited once by the read pointer. Justify O(1) space: only two integer pointers are used, no additional data structures.
Mention edge cases: empty array (return 0), all duplicates (return 1), no duplicates (return n). Discuss why in-place is beneficial (memory efficiency) and any potential drawbacks (modifies input).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the previous solution likely removed duplicates entirely (e.g., in a sorted array or linked list). Then explain that the modification requires allowing up to two occurrences of each value, which means adjusting the removal condition to permit one duplicate. Walk through the specific code changes, such as using a counter or a two-pointer technique with a gap of two, and justify why the time and space complexity remain optimal.
Pro tip: Emphasize that the key is to change the condition from 'if current != previous' to 'if current != previous or count < 2', and mention that this pattern generalizes to 'at most k duplicates' by tracking a count or using a pointer that lags by k. This shows you understand the underlying principle, not just the specific case.
Briefly summarize the original problem (e.g., remove duplicates from sorted array) and the approach used (e.g., two-pointer technique). This sets the context for the modification.
Explain that instead of allowing each value exactly once, we now allow at most twice. This means we need to permit one duplicate for each value.
Modify the condition that decides whether to keep the current element. For example, in the two-pointer approach, compare the current element with the element at write_index - 2 (if it exists) instead of write_index - 1. If they are different, it's safe to write; otherwise, skip.
Trace the modified algorithm on a sample input (e.g., [1,1,1,2,2,3]) to show how it correctly keeps at most two of each value and returns the new length.
State that time complexity remains O(n) and space O(1). Discuss edge cases like arrays with length 0, 1, or 2, and all elements the same.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.