Classic majority vote problem extended to two candidates instead of one.
Start by clarifying the problem constraints and edge cases, then discuss the Boyer-Moore Majority Vote algorithm as the optimal solution. Explain that at most two elements can appear more than n/3 times, and use two candidates and counters to find them in O(n) time and O(1) space, followed by a verification pass.
Pro tip: Mention that the verification pass is crucial because the Boyer-Moore algorithm only guarantees candidates, not actual majorities. Also, note that using a hash map is a valid alternative but may not meet the O(1) space requirement if that's expected.
Ask about input size, whether the array can be empty, and if the output order matters. Confirm that 'more than n/3 times' means strictly greater than floor(n/3).
Mention brute force (O(n^2)), hash map (O(n) time, O(n) space), and the optimal Boyer-Moore Majority Vote (O(n) time, O(1) space). Explain why the optimal is preferred.
Describe maintaining two candidates and two counters. Iterate through the array: if current equals a candidate, increment its counter; else if a counter is zero, set candidate; else decrement both counters.
After the first pass, do a second pass to count the occurrences of each candidate. Include only those with count > n/3 in the result.
State time O(n) and space O(1). Walk through a small example to demonstrate correctness and handle edge cases like empty array or no majority.
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