My first instinct was a nested loop and I almost went with it before catching myself.
Start by clarifying that generating 10,000 Bernoulli samples is inherently O(n²) since each of the n² entries must be generated. Then focus on the normalization step: compute column sums in O(n²) by iterating over all elements, and divide each element by its column sum in another O(n²) pass. Emphasize that this two-pass approach is optimal because you must touch every element at least once.
Pro tip: Mention that you can combine the two passes into one by first computing column sums, then normalizing in-place, but be prepared to discuss the trade-off between memory and cache efficiency. Also, note that using vectorized operations (e.g., NumPy) is practically faster but still O(n²) in complexity.
Confirm that the matrix is 100x100, each entry is a Bernoulli(0.5) sample (0 or 1), and normalization means each column sums to 1. Discuss that since entries are 0/1, column sums are counts of ones, and normalization divides each entry by the column sum.
Describe generating the matrix with nested loops (O(n²)) and then normalizing by computing column sums and dividing each element. This is straightforward but may involve multiple passes.
Explain that the optimal approach is to compute column sums in one pass over the matrix (O(n²)), then divide each element by its column sum in a second pass (O(n²)). This is O(n²) overall and optimal because every element must be read and written.
Mention in-place normalization to save memory, handling columns with sum zero (though unlikely with p=0.5 and n=100), and using vectorized operations for practical speed. Note that the algorithm is O(n²) regardless of vectorization.
Write clear pseudocode or actual code (e.g., in Python with NumPy) demonstrating the two-pass approach, ensuring it runs in O(n²) time and O(n²) space (or O(n) extra space for column sums).
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