Model the problem as a graph where products are nodes and pairs are edges, then find connected components using Union-Find (Disjoint Set Union) for near-constant time operations. The number of components gives the number of categories, and the size of each component gives the category sizes.
Pro tip: Mention that Union-Find with path compression and union by rank is optimal for dynamic connectivity and scales well for large inputs, which is crucial for Amazon-scale data.
Confirm that pairs are undirected, transitivity applies, and handle cases like empty input, self-pairs, or duplicate pairs. Ask if products are identified by integers or strings.
Select Union-Find (Disjoint Set Union) to efficiently group products into categories. Alternatively, consider BFS/DFS on an adjacency list if the graph is sparse and static.
Initialize each product as its own parent. For each pair, union the two sets using union by rank/size and path compression to keep operations nearly O(1).
After processing all pairs, iterate through all products, find their root, and tally the size of each root. The number of distinct roots is the number of categories.
State time complexity: O(N + M α(N)) where N is number of products, M is number of pairs, and α is the inverse Ackermann function. Space: O(N). Compare with BFS/DFS which is O(N+M) but may use more memory for adjacency lists.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the courses and prerequisites as a directed graph, then perform a topological sort. If a cycle exists, not all courses can be completed; otherwise, the topological order gives a valid sequence.
Pro tip: Clarify edge direction (prerequisite -> course) upfront and mention that Kahn's algorithm naturally detects cycles when the processed count is less than n. This shows attention to detail and algorithmic insight.
Represent each course as a node and each prerequisite pair (a, b) as a directed edge from a to b, meaning a must be taken before b.
Select either Kahn's algorithm (BFS with in-degree) or DFS with cycle detection. Explain the trade-offs briefly.
Run the algorithm, tracking visited nodes and detecting back edges (DFS) or nodes with non-zero in-degree (Kahn's). If a cycle is found, return false.
If no cycle, return the topological order as a valid course sequence. Otherwise, explain that a cycle prevents completion.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a prefix sum approach: iterate through the array while maintaining a running sum, and check if the current sum has been seen before. If it has, the subarray between the previous occurrence and the current index sums to zero. This yields an O(n) time and O(n) space solution.
Pro tip: Mention that this problem is equivalent to finding two equal prefix sums, and that the same technique extends to finding a subarray with sum equal to any target K. Also, note that if the array contains a zero, the answer is trivially true, but the prefix sum method handles it naturally.
Confirm that the subarray must be non-empty and contiguous, and that the array can contain negative numbers, zeros, and duplicates. Ask if the array can be empty or if there are any constraints on size.
Define prefix sum at index i as the sum of elements from 0 to i. A subarray from i+1 to j sums to zero if and only if prefix sum at j equals prefix sum at i.
Initialize a hash set with 0 to handle subarrays starting at index 0. Iterate through the array, updating the running sum. If the sum is already in the set, return true; otherwise, add it to the set. If the loop completes, return false.
State that the time complexity is O(n) because each element is processed once, and space complexity is O(n) in the worst case for the hash set.
Mention edge cases: empty array (return false), array with a zero (return true), and all positive numbers (return false). Briefly note that a brute-force O(n^2) approach exists but is inefficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Sliding window with a frequency counter diff.
Use a sliding window of the last len(target) characters, maintaining a frequency count of characters in the window and comparing it to the target's frequency count. To avoid O(k) comparison per step, track the number of matching characters or use a rolling hash of frequencies. Report the current position whenever the window's frequency map matches the target's.
Pro tip: Mention that you can optimize the comparison by maintaining a count of characters that have the correct frequency, reducing the per-step check to O(1). Also, clarify that the stream is processed online, so you only store the window and frequency maps, not the entire stream.
Confirm that the stream is infinite, characters arrive one at a time, and we need to report positions where the last len(target) characters form an anagram. Discuss memory constraints and whether the alphabet is fixed (e.g., ASCII).
Maintain a frequency map for the target and for the current window of size len(target). Use a queue or circular buffer to efficiently add new characters and remove old ones as the window slides.
Instead of comparing full frequency maps each step, track the number of characters whose frequencies match between the window and target. Update this count incrementally when adding or removing a character.
For each incoming character, update the window and the match count. When the window size equals len(target) and the match count equals the number of distinct characters in the target, report the current position.
State that time complexity is O(n) for n characters, with O(1) per character (assuming fixed alphabet). Space complexity is O(k) for the frequency maps, where k is the alphabet size. Discuss alternative approaches like rolling hash and their trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.