My first instinct was a hashmap to count frequencies, which works fine and I got it out cleanly.
Start by clarifying the problem and constraints, then propose a solution using Boyer-Moore Voting Algorithm, which finds the majority element in O(n) time and O(1) space. Explain the algorithm's intuition and walk through a small example to demonstrate correctness.
Pro tip: Mention that the Boyer-Moore algorithm is optimal for this problem and that it can be extended to find elements appearing more than n/k times. Also, discuss potential edge cases like when the majority element is not guaranteed, and how to verify the candidate in a second pass.
Restate the problem to ensure understanding: find the element that appears more than n/2 times in a non-empty array where a majority element always exists. Ask if there are any constraints on time or space complexity.
Mention brute force (O(n^2)), sorting (O(n log n)), hash map (O(n) time, O(n) space), and then introduce Boyer-Moore Voting Algorithm as the optimal O(n) time, O(1) space solution.
Describe the algorithm: maintain a candidate and a count; iterate through the array, if count is 0 set candidate to current element and count to 1; else if current element equals candidate increment count, else decrement count. The candidate at the end is the majority element.
Choose a small array (e.g., [2,2,1,1,1,2,2]) and step through the algorithm to show how the candidate and count evolve, demonstrating that the final candidate is the majority element.
State that the algorithm runs in O(n) time and O(1) space. Discuss that if the majority element is not guaranteed, a second pass is needed to verify the candidate's count. Also mention that the algorithm works because the majority element's count cannot be canceled out completely.
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