Knew this was a sliding window problem pretty fast, which was a relief.
Use a sliding window with two pointers to expand and contract a window over s while tracking character counts of t. Maintain a count of matched characters to know when the window is valid, and record the smallest valid window. This yields O(n) time and O(k) space, where k is the number of distinct characters in t.
Pro tip: Clarify edge cases upfront (e.g., t longer than s, empty strings, Unicode) and mention that the algorithm runs in O(n) time, which is optimal since you must scan s at least once. Also, discuss how you would handle large inputs or streaming data if relevant.
Restate the problem to confirm requirements: smallest substring of s containing all characters of t, including duplicates. Ask about edge cases: empty strings, t longer than s, characters outside ASCII, and whether the answer is guaranteed to exist.
Propose a sliding window (two-pointer) technique because it efficiently finds the minimum window in linear time. Explain why brute force is inefficient and why sliding window is optimal.
Use a hash map to count characters in t. Expand the right pointer to include characters, updating counts and a 'formed' counter. When all characters are matched, contract the left pointer to minimize the window while keeping it valid, updating the minimum length and start index.
State time complexity O(n) and space O(k) where k is distinct characters in t. Walk through a small example to verify correctness, and consider edge cases like no valid window.
Mention possible optimizations (e.g., filtering s to only relevant characters) and variations (e.g., finding all minimum windows, handling streaming input).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the exclusion rules as an undirected graph where each animal is a vertex and each exclusion is an edge. The rules are conflict-free if and only if the graph is bipartite, meaning it can be 2-colored without adjacent vertices sharing the same color. Use BFS or DFS to check bipartiteness, handling disconnected components.
Pro tip: Clarify that 'conflict-free' means no two animals that can't be together are forced to be on the same bank. Mention that self-loops (an animal excluding itself) immediately make it impossible, and that the problem reduces to bipartite checking.
Restate the problem: we need to assign each animal to one of two banks such that no exclusion rule is violated. Recognize that this is equivalent to 2-coloring a graph.
Create vertices for each animal. For each key and each animal in its exclusion list, add an undirected edge between them. Ensure the graph is undirected and handle duplicates.
Use BFS or DFS to attempt to 2-color the graph. Start from each unvisited vertex, assign a color, and propagate colors to neighbors, flipping the color at each step. If a neighbor already has the same color, a conflict exists.
Check for self-loops (an animal excluding itself) which immediately cause a conflict. Also consider disconnected components and ensure all vertices are visited.
If all components are successfully 2-colored without conflicts, return true (conflict-free). Otherwise, return false.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.