Recognized it as the LeetCode reaching points problem pretty fast, which was a relief.
Work backwards from (c, d) to (a, b) by repeatedly subtracting the smaller number from the larger, since the forward operations only increase one coordinate. Use modulo arithmetic to skip many steps at once, similar to the Euclidean algorithm, and handle edge cases like zero or negative numbers.
Pro tip: Mention that this is essentially the reverse of the subtractive Euclidean algorithm, and that using modulo instead of repeated subtraction reduces time complexity from O(max(c,d)) to O(log(max(c,d))). Also, clarify that the operations are only valid for positive integers, so if any coordinate is non-positive, the answer is immediately false.
Recognize that each operation adds one coordinate to the other, so both coordinates remain positive and strictly increase. Thus, if (a, b) is not component-wise ≤ (c, d), it's impossible.
From (c, d), the last operation must have been either (c-d, d) if c > d, or (c, d-c) if d > c. So we can reverse by subtracting the smaller from the larger.
Instead of subtracting one at a time, compute how many times the smaller fits into the larger: if c > d, replace c with c % d, but ensure we don't skip past the target. Use a while loop until one coordinate equals the corresponding target coordinate.
After reducing, check if (a, b) is reached exactly. If at any point a coordinate becomes ≤ 0 or the reduction overshoots, return false. Also handle the case where one coordinate is zero: if a=0 or b=0, then (c,d) must be a multiple of the other coordinate.
The modulo-based approach runs in O(log(max(c,d))) time. Discuss edge cases: a or b zero, negative inputs, and the case where (a,b) equals (c,d) initially.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one caught me off guard because it looks like a simple greedy pairing problem but there are edge cases around tasks of the same type that still can't be paired due to memory.
Group tasks by type, then for each type, sort memory requirements and use a greedy two-pointer approach to pair the largest with the smallest possible task that fits within the limit. Count the number of pairs (each pair takes 1 unit of time) and add the number of unpaired tasks to get the total time.
Pro tip: Clarify that tasks of different types cannot run in parallel, so the total time is the sum of times for each type. Also, mention that the greedy pairing is optimal because pairing the largest with the smallest maximizes the chance of fitting two tasks together, reducing the total time.
Create a mapping from each task type to a list of memory requirements for tasks of that type.
For each type, sort the list of memory requirements in ascending order to facilitate efficient pairing.
Use two pointers (left at smallest, right at largest) to pair the largest remaining task with the smallest possible task that fits within the memory limit. If a pair fits, increment the pair count and move both pointers; otherwise, move the right pointer left (the largest task must run alone).
For each type, the time required is the number of pairs plus the number of unpaired tasks (which equals the total number of tasks minus the number of pairs).
Add the times for all types to get the total minimum time to complete all tasks.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.