Choose an approach without the label

What is a vertex, and what quantity should order the unfinished work?

The floor is sufficient state under fixed button rules. BFS solves fewest presses; there are 31 vertices and at most two outgoing moves each. The first answer is six presses: five ups and one down, ordered to stay in bounds. With different nonnegative time costs, Dijkstra is a valid general approach. In this particular input the same move counts also minimize time, but BFS’s usual proof does not establish that for arbitrary weighted transitions. See BFS and Dijkstra.

What remains after taking the current session? What did a zero base case permit?

Compare skip best(i+1) with take benefit[i]+best(i+2). One suffix index suffices, with O(n) states and constant work each. The first result is 8. For the nonempty negative case, the answer is -2; the original empty-allowed recurrence would incorrectly return zero. Enforce nonempty feasibility through state or an appropriate reformulation. See 1D DP.

Does the original board location determine legal next moves? Does movement impose a DAG?

Use BFS over (row,column,budget_left), with discovered state including the budget. The expanded graph has O(RC(K+1)) states and bounded outdegree for general budget K. Four-direction cycles invalidate the unchanged right/down DP recurrence. More remaining budget can sometimes support a carefully proved dominance optimization, but do not discard that coordinate without one. See state and budget DP.

Which branch information should go upward, and which two-branch candidate can you measure at a node?

Root the tree, compute downward heights, and combine the best two child branches at every node. Retain the maximum candidate across nodes. O(V) time; stack O(h). In an adjacency-list tree, skip the parent edge. Returning a diameter as though it were a single extendable height is a common wrong combination. See diameter.

What relationship does each query use, and how much output must it produce?

A trie shares prefix work and then traverses a matching subtree. A set may be sufficient for the exact-membership workload. Neither bypasses string hashing/comparison or output costs in a detailed model. A trie end marker distinguishes a word from a prefix. See tries and autocomplete.

Readiness supplies an order; what summary travels along dependency edges?

Topologically process ready tasks. Earliest finish of a task is its duration plus the maximum predecessor finish, with zero for no predecessor. The total finish is the maximum task finish. Check that every task was processed to detect cycles. Time O(V+E). Summing every duration ignores parallelism. See dependencies.

Separate availability from remaining capacity, and define the empty suffix’s answer.

Use ways(i,remaining), adding skip and feasible take, both advancing the item index. With no items left, return one only if remaining is zero. A negative remaining capacity is impossible. State bound O(nW), constant transitions each. Value-based deduplication would remove required physical-item choices. See knapsack and coin semantics.

How many reading positions are independent, and what does a mismatch mean?

The first is LCS over two suffix positions with matching or skipping transitions, O(mn) time and space in the basic implementation. The contiguous version needs a different contract, such as matching length ending at (i,j), resetting to zero on mismatch. A two-coordinate shape does not uniquely determine the recurrence. See string DP.

Are you paying once for a network, or measuring separate routes?

The first calls for an MST when the graph is connected: Kruskal plus disjoint sets is one approach, dominated by sorting edges. The second is a shortest-path problem; Dijkstra fits nonnegative costs. The triangle with costs 2,2,3 distinguishes the results. An MST need not preserve cheapest individual routes. See spanning trees.

Can adding another day repair an over-limit sequence?

With nonnegative expenses and limit, a moving sum window has safe monotonic updates and takes O(n) time. Refunds break that justification: [6,-2] under limit 5 is a counterexample. Re-derive a prefix-based approach or use a quadratic baseline for small inputs; the exact-sum frequency trick solves a different objective and cannot be copied unchanged. See windows.

What choice at the boundary changes which child answer is valid?

Return taken/skipped values for each subtree and combine conditionally, O(n) time on a tree. Cross-team edges can couple formerly independent branches, so the same recurrence can become wrong. Do not assume a two-value summary still describes those interactions. See tree DP.

Which structure shares possible spellings, and which memory belongs only to the active route?

A trie guides a backtracking board search; a path-local visited set prevents repeated cells. A set of found words satisfies the first contract but loses required coordinate paths in the second. Worst-case search remains exponential in useful word length, and output size matters. See board word search.