Classic sliding window but the multiplicity part tripped me up at first.
Use a sliding window (two-pointer) technique to find the shortest substring of s that contains all characters of t with the required frequencies. Expand the right pointer to include characters until the window is valid, then shrink the left pointer to find the minimal valid window. Track the minimum length and starting index, returning the substring or an empty string if no valid window exists.
Pro tip: Clarify edge cases upfront (e.g., t longer than s, empty strings) and discuss how your solution handles them. Also, mention that the sliding window approach is optimal because it avoids re-scanning characters, achieving O(n) time.
Restate the problem to ensure clarity: find the shortest contiguous substring of s that contains every character in t with at least the same frequency. Discuss edge cases: if t is empty, return empty string; if s is shorter than t, return empty string; if no such substring exists, return empty string.
Use a frequency map (e.g., hash map or array of size 128 for ASCII) to count characters in t. Maintain a window frequency map and a counter for how many characters of t are satisfied (i.e., have the required frequency in the window).
Initialize left and right pointers at 0. Expand right to include characters, updating the window frequency and satisfied counter. When all characters are satisfied, shrink left to minimize the window, updating the minimum length and start index. Continue until right reaches the end of s.
After the loop, if a valid window was found, return the substring from the recorded start index with the minimum length. Otherwise, return an empty string.
State time complexity O(n) where n is the length of s, as each character is visited at most twice. Space complexity O(k) where k is the number of unique characters in t (or O(1) if using fixed-size array). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as interval graph coloring: sort jobs by start time, use a min-heap to track the earliest finishing job among active workers, and assign a new worker only when no existing worker is free. Then map the assignments back to the original input order.
Pro tip: Emphasize that the greedy algorithm is optimal because the maximum number of overlapping jobs at any point is a lower bound, and the algorithm achieves exactly that many workers. Also, clarify that half-open intervals mean a job ending at time t does not conflict with one starting at t.
Confirm that intervals are half-open [start, start+duration) and that jobs are independent. Model the problem as interval graph coloring where colors represent workers.
Create events for each job: (start, +1) and (end, -1). Sort events by time, processing starts before ends to respect half-open intervals. Track the maximum number of concurrent jobs to determine the minimum workers needed.
Sort jobs by start time. Use a min-heap of (end_time, worker_id) for active workers. For each job, if the earliest end_time <= start, reuse that worker; otherwise, assign a new worker. Push the job's end_time and worker_id onto the heap.
Store the worker assignment for each job in an array indexed by the job's original position. Return this array as the result.
State that sorting takes O(n log n) and heap operations take O(n log n), so overall O(n log n) time and O(n) space. Argue optimality by noting that the maximum overlap is a lower bound and the greedy uses exactly that many workers.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.