I knew the three-pointer approach the second I read it, but explaining the invariants out loud while coding was trickier than expected.
Use the Dutch National Flag algorithm with three pointers (low, mid, high) to partition the array in a single pass. Iterate mid from start to high, swapping elements to place 0s before low, 2s after high, and leaving 1s in the middle. This achieves O(n) time and O(1) space.
Pro tip: Emphasize that this is a classic three-way partitioning problem and that the algorithm is optimal because it sorts in one pass without extra space. Mention that it's also known as the Dutch National Flag problem, which shows depth of knowledge.
Confirm that the array contains only 0s, 1s, and 2s, and that sorting must be in-place with O(n) time and O(1) extra space. Ask if stability matters (it usually doesn't for this problem).
Name the Dutch National Flag algorithm and explain the three-pointer approach: low, mid, and high. Describe the invariant: elements before low are 0s, between low and mid are 1s, after high are 2s, and the unknown region is between mid and high.
Explain the loop: while mid <= high, if arr[mid] == 0, swap arr[low] and arr[mid], increment low and mid; if arr[mid] == 1, increment mid; if arr[mid] == 2, swap arr[mid] and arr[high], decrement high. Show how this maintains the invariant.
State that each element is examined at most once, so time complexity is O(n). Space complexity is O(1) because only a constant number of pointers are used.
Walk through a small example like [2,0,1,2,0,1] to demonstrate correctness. Mention edge cases: all same elements, already sorted, reverse sorted, empty array.
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