Solve capital selection and grid escape

Implement two functions for algorithmic practice:

1. MaximizeCapitalAfterProjects(k, initialCapital, requiredCapital, profit): You may pick at most k projects. At each selection, you can start any project whose requiredCapital[i] <= current capital and immediately gain profit[i]. Return the maximum capital achievable after up to k selections. Optimize for n up to 2e5 and justify data-structure choices and time/space complexity.
2. MinTimeToExitSewer(grid, heights): grid contains 'S' (start), 'E' (exit), '#' (wall), and '.' (open). heights is an equally sized matrix of nonnegative integers. You move 4-directionally one cell per minute, starting at time 0 from 'S'. At minute t, all cells with heights[r][c] <= t are flooded and impassable. Find the minimum arrival time to 'E' before flooding, or return -1 if impossible. Discuss algorithm, correctness, edge cases, and complexity for grids up to 5e5 cells.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Maximize Capital After Projects

You are given an integer k (the maximum number of projects you may complete), an integer initialCapital, and two integer arrays requiredCapital and profit of equal length n. Project i requires requiredCapital[i] capital on hand to start, and completing it adds profit[i] to your capital (the required capital is NOT consumed — it is only a threshold).

Starting from initialCapital, you may complete at most k projects, one at a time. Before each pick, every project whose requiredCapital[i] <= current capital is startable; you choose one, add its profit to your capital, and that project cannot be reused. Choosing greedily, return the maximum capital achievable after at most k selections.

This is the classic "IPO" problem. The optimal strategy at every step is to take the most profitable project that is currently affordable. Sort projects by required capital so you can unlock newly-affordable projects as capital grows, and keep the profits of all currently-affordable projects in a max-heap. Each of up to k rounds pops the largest available profit.

Return initialCapital early if no project is ever affordable or k == 0.

Constraints

• 0 <= k <= n
• 1 <= n <= 2 * 10^5
• 0 <= initialCapital <= 10^9
• 0 <= requiredCapital[i] <= 10^9
• 0 <= profit[i] <= 10^4
• requiredCapital.length == profit.length

Examples

Example 1

Input: (2, 0, [0, 1, 1], [1, 2, 3])
Output: 4

Explanation: Start cap=0. Only project 0 (req 0, profit 1) is affordable; take it -> cap=1. Now projects 1,2 (req 1) are affordable with profits 2,3; take 3 -> cap=4. After 2 picks, max capital is 4.

Example 2

Input: (3, 0, [0, 1, 2], [1, 2, 3])
Output: 6

Explanation: Pick profit 1 (cap 0->1), then profit 2 (cap 1->3), then profit 3 (req 2<=3, cap 3->6). All three picked greedily by affordability.

Hints

1. At each step the optimal choice is the most profitable project you can currently afford — a greedy argument: taking a smaller profit now never increases what you can reach later.
2. Sort projects by required capital, then advance an index to unlock every project whose threshold is now <= your capital, pushing their profits into a max-heap.
3. Use a max-heap (negate values with Python's min-heap). Each round pops the largest affordable profit; stop early if the heap is empty (nothing affordable) or after k picks.

Minimum Time to Exit the Flooding Sewer

You are given a character grid grid and an equally sized integer matrix heights of nonnegative values. The grid contains exactly one 'S' (start), exactly one 'E' (exit), '#' cells (walls, impassable), and '.' cells (open). You move 4-directionally (up/down/left/right), one cell per minute, starting at time 0 on the 'S' cell.

Flooding rule: at minute t, every cell whose heights[r][c] <= t is flooded and impassable. You may occupy a cell at minute t only if it is NOT flooded at that minute, i.e. heights[r][c] > t. In particular, the start must satisfy heights[start] > 0, otherwise you can never even stand on it.

Return the minimum arrival time at 'E', or -1 if it is impossible to reach the exit before it floods (or if there is no path at all).

Because every move costs exactly one minute, the first time BFS reaches a cell is the earliest possible arrival time. Run a standard BFS from 'S': when stepping into a neighbor at arrival time nt = t + 1, reject it if it is a wall or if it would already be flooded (heights <= nt). Edge cases: missing S/E, S == E (answer 0 when heights[S] > 0), start cell already flooded, and an exit walled off or always flooded.

Constraints

• 1 <= R * C <= 5 * 10^5
• grid[r][c] is one of 'S', 'E', '#', '.'
• Exactly one 'S' and at most one 'E'
• 0 <= heights[r][c] <= 10^9
• heights has the same dimensions as grid

Examples

Example 1

Input: ([['S', '.', 'E']], [[9, 9, 9]])
Output: 2

Explanation: Straight corridor with tall heights. Move S(t0)->.(t1)->E(t2). Earliest arrival is 2.

Example 2

Input: ([['S', '#', 'E']], [[9, 9, 9]])
Output: -1

Explanation: A wall '#' separates S from E and there is no alternate route, so E is unreachable.

Hints

1. Every move costs exactly one minute, so a uniform-cost BFS from S labels each reachable cell with its earliest possible arrival time — the first time you touch a cell is optimal.
2. When stepping into a neighbor that you would arrive at time nt = t + 1, it must NOT be flooded yet: require heights[neighbor] > nt. Walls ('#') are always blocked.
3. Handle the corner cases first: no S or no E -> -1; the start itself flooded (heights[S] <= 0) -> -1; S already on E -> 0.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Implement two functions for algorithmic practice: 1) MaximizeCapitalAfterProjects(k, initialCapital, requiredCapital, profit): You may pick at most k projects. At each selection, you can start any project whose requiredCapital[i] <= current capital and immediately gain profit[i]. Return the maximum capital achievable after up to k selections. Optimize for n up to 2e5 and justify data-structure choices and time/space complexity. 2) MinTimeToExitSewer(grid, heights): grid contains 'S' (start), 'E' (exit), '#' (wall), and '.' (open). heights is an equally sized matrix of nonnegative integers. You move 4-directionally one cell per minute, starting at time 0 from 'S'. At minute t, all cells with ...

Recommended Approach

Represent each cell as a state. Use BFS for minimum-distance propagation, DFS with memoization for longest monotonic paths, and careful boundary checks for simulation. Store obstacles or visited cells in sets when the grid is sparse.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

A full grid traversal is O(mn) time and O(mn) space in the worst case. Sparse simulation is O(number_of_commands) plus obstacle storage.

Edge Cases and Tests

Empty grid, one row/column, blocked start or target, boundaries, repeated visits, and tie-breaking among equal-distance cells.