← MathWorks Interview Insights
I knew there was a greedy angle here but took me a bit to convince myself why merging the two smallest first actually works.
Recognize that the problem is equivalent to building a Huffman tree, where the total cost is the sum of all internal node weights. The optimal strategy is to always combine the two smallest numbers first, which minimizes the total cost. Explain this using a greedy algorithm and justify why it works.
Pro tip: Mention that this is exactly the Huffman coding algorithm and that using a min-heap gives an efficient O(n log n) solution. Also, briefly explain why greedy works here (exchange argument) to show depth.
Restate the problem: we need to merge numbers pairwise, paying the sum each time, and minimize the total cost. Recognize that the order of merges affects the total cost.
Propose the greedy strategy: always pick the two smallest numbers to merge. Explain that this minimizes the immediate cost and leads to the global minimum.
Explain that this is exactly the Huffman coding problem, where the total cost is the weighted external path length. The greedy algorithm builds the optimal Huffman tree.
Provide a brief justification: by exchange argument, any optimal solution must merge the two smallest first; otherwise swapping would reduce cost. This ensures the greedy choice is safe.
Describe using a min-heap (priority queue) to efficiently extract the two smallest and insert their sum. Time complexity is O(n log n), space O(n).
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