Use the Dutch National Flag algorithm with three pointers (low, mid, high) to partition the array in a single pass. Start with low and mid at the beginning, high at the end, and iterate while mid <= high, swapping elements to group 0s, 1s, and 2s. This achieves O(n) time and O(1) space, which is optimal for this problem.
Pro tip: Mention that this is a classic three-way partitioning problem and that the Dutch National Flag algorithm is the optimal solution. Also, discuss edge cases like empty arrays or arrays with only one color, and emphasize that the solution is in-place and stable is not required.
Confirm that the array contains only 0, 1, and 2, and that sorting must be in-place. Ask if stability is required (usually not) and if the array can be modified.
Select the Dutch National Flag algorithm for O(n) time and O(1) space. Explain why it's better than counting sort (which requires two passes) or general sorting (O(n log n)).
Describe the three pointers: low (boundary for 0s), mid (current element), high (boundary for 2s). Explain the swap rules: if arr[mid] == 0, swap with low and increment both; if 1, increment mid; if 2, swap with high and decrement high.
State time complexity O(n) and space O(1). Discuss edge cases: empty array, all elements same, already sorted, and arrays with only two colors.
Trace through a small example like [2,0,2,1,1,0] to demonstrate correctness and show how the pointers move.
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