I'd seen this before so the cycle detection angle clicked fast.
Recognize this as the classic 'happy number' problem. Explain that the sequence either reaches 1 or enters a cycle, and propose using a hash set to detect cycles or Floyd's cycle-finding algorithm for O(1) space. Then discuss the time complexity and why numbers cannot grow indefinitely.
Pro tip: Mention that for any number with more than 3 digits, the sum of squares of digits is strictly less than the number itself, so the sequence is bounded and must eventually cycle or reach 1. This shows deep insight and can lead to a more efficient solution.
Restate the problem: given a positive integer, repeatedly replace it with the sum of the squares of its digits. Determine if it eventually reaches 1 or enters a cycle without 1.
Explain that the sequence either reaches 1 (happy number) or enters a cycle that does not include 1 (unhappy number). The cycle for unhappy numbers is known to be 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.
Propose using a hash set to store seen numbers and detect repeats, or use Floyd's tortoise and hare algorithm for O(1) space. Discuss trade-offs.
Discuss time complexity: O(log n) per step, and the number of steps is bounded. Space complexity: O(1) with Floyd's, O(k) with hash set. Handle edge cases like n=1 (immediately happy) and n=0 (not positive, but if allowed, 0 loops).
Write clean code for the chosen approach, and test with examples like 19 (happy) and 2 (unhappy). Mention that the cycle for unhappy numbers is fixed, so you can also just check if the number becomes 4.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Base case was straightforward BFS, nothing surprising.
Start by modeling the grid as a graph and use BFS for the shortest path when all cells are open. For the extension with up to k blocked cells, use 0-1 BFS or Dijkstra with state (row, col, remaining_k) to track the minimum number of blocked cells broken. Discuss trade-offs between time/space complexity and optimality.
Pro tip: Mention that 0-1 BFS is optimal for the extension because edge weights are 0 or 1, and it runs in O(mn) time, which is better than Dijkstra's O(mn log(mn)). Also, clarify that the path length is measured in steps, not number of blocked cells converted.
Ask about grid size, movement directions (4-way or 8-way), and whether k is fixed or variable. Confirm that the goal is to minimize path length, not the number of conversions.
Explain that BFS finds the shortest path in an unweighted grid. Describe the algorithm: queue, visited set, and level-by-level traversal.
Introduce state (r, c, rem) where rem is remaining conversions. Use 0-1 BFS: moving to an open cell costs 0, moving to a blocked cell costs 1 (if rem > 0). Alternatively, use Dijkstra with a priority queue.
Compare BFS (O(mn)) for base case vs. 0-1 BFS (O(mn)) for extension. Discuss space complexity O(mn * k) if using a 3D visited array, but can be optimized to O(mn) by tracking minimum conversions per cell.
Consider cases where start or end is blocked, k=0, k is large, or no path exists. Walk through a small example to validate the approach.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.