Sorting and greedily picking from smallest works cleanly when everything is non-negative, but I stumbled when the interviewer brought up negatives.
Start by clarifying the problem and edge cases, then propose a greedy strategy: sort the array and pick the smallest elements first to maximize count. Discuss how zeros and negatives affect the greedy choice and complexity, and analyze time/space trade-offs.
Pro tip: Mention that zeros can be included for free and negatives should always be included as they increase the budget; this shows attention to edge cases and practical optimization.
Confirm that elements can be picked in any order, each at most once, and that the goal is to maximize the count while keeping the sum ≤ n. Ask about constraints (e.g., array size, value ranges) to inform algorithm choice.
Discuss zeros (they can always be included without affecting the sum) and negatives (they reduce the sum, so always include them). Also consider empty array, all positives, and n negative.
Sort the array in ascending order. Iterate through the sorted array, adding elements to the sum as long as the sum does not exceed n. Count the number of elements picked.
Time complexity: O(m log m) due to sorting, where m is the array length; space complexity: O(1) if sorting in-place, or O(m) if using extra space. Mention that if the array is already sorted or can be sorted in linear time (e.g., counting sort for bounded integers), time can be O(m).
Compare with other approaches (e.g., dynamic programming for subset sum) and explain why greedy is optimal here. Mention that if the array is huge and n is small, a min-heap could be used to avoid full sort, but sorting is generally efficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sort once, build a prefix sum array, then binary search for each query.
Start by clarifying the query semantics: what exactly is n and what output is expected? Then propose a preprocessing strategy such as sorting the array and building prefix sums or a segment tree, depending on the query type. Emphasize that preprocessing enables O(1) or O(log n) per query, which is crucial for 100,000 queries.
Pro tip: Mention that you would validate the preprocessing with a few sample queries and consider memory constraints, as Amazon often values scalable and production-ready solutions.
Ask what n represents (e.g., index, threshold, count) and what the query should return (e.g., sum, count, k-th element). This determines the preprocessing needed.
Based on the query type, select an appropriate data structure: prefix sums for range sums, sorted array with binary search for threshold queries, or segment tree for dynamic range queries.
Explain that preprocessing takes O(n log n) or O(n) time and O(n) space, while each query becomes O(1) or O(log n), making 100,000 queries efficient.
Discuss handling duplicate values, out-of-range n, and memory limits. Mention that if the array is static, we can precompute answers for all possible n if the range is small.
Propose testing with sample queries and considering trade-offs between preprocessing time and query time, especially if the number of queries is very large.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a probabilistic counting algorithm like Morris counter that processes each element in O(1) space and provides an estimate with bounded relative error. For a count with at most ±1 error, a simple approach is to maintain a counter and increment it with probability 1/(2^i) when the counter is i, then adjust the final estimate. Justify that the expected value of the estimate equals the true count and the variance is small, ensuring the error bound.
Pro tip: Mention that for exact counts with ±1 error, you can use a deterministic algorithm like Misra-Gries for frequent items, but for total count, a randomized algorithm like Morris counter is standard. Also, note that the error bound is probabilistic, not worst-case, and discuss trade-offs.
Confirm that the stream is too large to store and that we need an approximate count with at most ±1 error. Discuss whether the error is absolute or relative.
Select a space-efficient algorithm such as Morris counter for approximate counting. Explain its basic idea: maintain a counter i, and increment it with probability 1/(2^i) when an element arrives.
Detail the steps: initialize counter to 0; for each element, if random() < 1/(2^counter), increment counter. At the end, estimate count as 2^counter - 1.
Show that the expected value of the estimate equals the true count, and the variance is bounded, so with high probability the error is within a small factor. For ±1 error, note that Morris counter gives relative error, so for small counts it may be exact, but for large counts the error can be larger; thus, for absolute ±1 error, a different approach like maintaining a exact count up to a threshold might be needed.
Mention that Morris counter uses O(log log n) space, which is extremely efficient, but the error is probabilistic. For deterministic ±1 error, one could use a simple counter if the count fits in memory, but if not, no deterministic algorithm can guarantee ±1 error in sublinear space for arbitrary streams.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Iterate i from 1 to sqrt(N), and for each divisor i, collect i and N/i as factors, being careful to add the square root only once when N is a perfect square. Then sort the factors and return the Kth smallest, or -1 if there are fewer than K factors.
Pro tip: Mention that you can avoid sorting by collecting factors in two lists (small and large) and merging them, but sorting is fine for N up to 10^12 since the number of divisors is at most ~6720. Also, note that early termination is possible if you only need the Kth factor and K is small.
Clarify that N can be up to 10^12, so O(N) iteration is infeasible. We need an O(sqrt(N)) approach. Also, handle the case where N has fewer than K factors.
Loop i from 1 to floor(sqrt(N)). If i divides N, then i and N/i are both factors. Add i to a list of smaller factors and N/i to a list of larger factors (or a single list).
If i == N/i (i.e., N is a perfect square), add only one instance of i to the factor list. This ensures no duplicate counting.
After the loop, combine the factors (if using two lists, the smaller factors are in increasing order and the larger factors are in decreasing order, so you can merge them). Sort if necessary, then check if the list has at least K elements. If yes, return the Kth element; else return -1.
Time complexity is O(sqrt(N)) for the loop plus O(D log D) for sorting, where D is the number of divisors (at most ~6720 for N ≤ 10^12). Space complexity is O(D) to store the factors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.