I knew topological sort was involved but fumbled the base case explanation for like two minutes.
Define dp[u][k] as the number of paths from s to u with exactly k edges, initialize dp[s][0] = 1, and process k from 0 to L-1, updating dp[v][k+1] += dp[u][k] for each edge u→v. Since the graph is a DAG, iterating k in increasing order ensures all contributions to dp[u][k] are computed before they are used.
Pro tip: Mention that the DP can be optimized to O(N) space by keeping only the previous k layer, and note that the DAG property guarantees no cycles, so the iteration order is valid without needing topological sorting.
Clearly state that dp[u][k] represents the number of distinct paths from source s to node u that use exactly k edges.
Set dp[s][0] = 1 (one empty path from s to itself) and dp[u][0] = 0 for all u ≠ s.
For each edge u→v and each k from 0 to L-1, add dp[u][k] to dp[v][k+1], effectively extending paths by one edge.
Iterate k from 0 to L-1 in increasing order; for each k, process all edges. This works because the graph is a DAG and paths of length k+1 depend only on paths of length k.
After filling the DP table, sum dp[u][L] over all nodes u (or return dp[t][L] for a specific target t) to get the total number of paths of length exactly L.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clearly stating the time and space complexity of your DP solution, explaining how you derived them from the state and transition. Then describe how rolling arrays can reduce space by keeping only the necessary previous states, and discuss the trade-offs involved.
Pro tip: Always relate the space optimization to the specific problem constraints and mention that while rolling arrays reduce space, they may sacrifice the ability to reconstruct the solution. This shows you consider practical implications beyond just complexity.
Clearly state the time and space complexity of your DP solution, e.g., O(n^2) time and O(n^2) space, and briefly explain why.
Describe the DP state definition and recurrence relation, highlighting which previous states are needed for the current computation.
Explain that if the recurrence only depends on a limited number of previous rows or states, you can use rolling arrays to store only those, reducing space complexity.
Show how to implement rolling arrays, e.g., using two rows instead of a full 2D table, and state the new space complexity, like O(n) or O(1).
Mention any trade-offs, such as increased code complexity or loss of ability to backtrack the solution, and when it's appropriate to use this optimization.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by restating the DP framework for unweighted graphs, then explain how to extend it to weighted edges by storing path weights instead of edge counts. Discuss the recurrence for exactly k edges and how to handle at most k edges, including the implications for shortest vs. longest paths.
Pro tip: Mention that for longest path with exactly k edges, the DP can be solved in polynomial time, but for at most k edges, it becomes NP-hard if k is part of the input due to the longest path problem. This shows awareness of complexity trade-offs.
Define dp[v][k] as the shortest (or longest) path weight from source to vertex v using exactly k edges. Initialize dp[source][0] = 0 and others to infinity (or -infinity for longest).
For each edge (u, v) with weight w, update dp[v][k] = min(dp[v][k], dp[u][k-1] + w) for shortest path (or max for longest). Iterate k from 1 to K.
For at most k edges, take the minimum (or maximum) over all dp[v][i] for i ≤ k. Alternatively, add a self-loop of weight 0 to allow 'wasting' edges, but careful with longest path.
Time complexity is O(K * (V + E)) for exactly k edges. Space can be optimized to O(V) by keeping only previous layer. Mention that for shortest path, this is equivalent to Bellman-Ford when K = V-1.
For longest path with exactly k edges, DP works in polynomial time. For at most k edges, it's NP-hard if k is part of input (reduction from Hamiltonian path). Clarify that if k is fixed, it's polynomial.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the original problem (likely counting paths from a source to all nodes in a DAG) and explain that to count paths to a specific target t, you can either compute the full DP and return only the value for t, or optimize by pruning branches that cannot reach t. Discuss trade-offs between simplicity and efficiency, and mention how to handle cycles if the graph is not a DAG.
Pro tip: Mention that if the graph is a DAG, you can compute paths to t by reversing edges and doing a DP from t, which naturally counts paths from all nodes to t and avoids unnecessary computation for nodes that cannot reach t.
Restate the original problem (e.g., counting paths from a source to all nodes in a DAG) and confirm assumptions like graph type, whether paths are simple, and if nodes can be revisited.
Explain that instead of returning a map or array of counts for all nodes, you only need the count for node t. This may allow early termination or pruning.
If the original DP computes counts for all nodes, simply run it and return the value for t. This is O(V+E) and simplest if the graph is small or if you need counts for other nodes anyway.
For efficiency, reverse the graph and compute paths from t to all nodes (or do a forward DP but only explore nodes that can reach t). This avoids computing counts for irrelevant nodes.
Compare time/space complexity of both approaches, and mention handling of cycles (e.g., using DFS with memoization and cycle detection, or noting that path counting in cyclic graphs may be infinite).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.