The two-table vs rolling-array distinction is where I fumbled a bit.
Start by defining the DP state and recurrence for the 0/1 knapsack, then implement the full O(nW) table to reconstruct the chosen items. After that, explain the space-optimized O(W) version, noting that reconstruction requires either storing parent pointers or re-running the DP with a copy of the previous row.
Pro tip: Emphasize that the space-optimized version cannot reconstruct the solution without additional storage; discuss trade-offs between storing decisions (O(nW) bits) or re-computing (O(nW) time). This shows you understand practical constraints beyond just the algorithm.
Confirm that each item can be taken at most once (0/1), and ask about input sizes, value ranges, and whether reconstruction is required. This helps decide between DP approaches.
Let dp[i][w] be the max value using first i items with capacity w. Recurrence: dp[i][w] = max(dp[i-1][w], dp[i-1][w-w_i] + v_i) if w_i <= w, else dp[i-1][w].
Fill the (n+1) x (W+1) table. To reconstruct, backtrack from dp[n][W]: if dp[i][w] != dp[i-1][w], item i was taken; update w -= w_i and i--. Otherwise, i--.
Use a 1D array dp[w] and iterate items, updating w from W down to w_i to avoid reusing items. For reconstruction, either keep a 2D boolean array of decisions or re-run the DP with a copy of the previous row.
Full DP: O(nW) time and space. Optimized: O(nW) time, O(W) space, but reconstruction needs extra O(nW) bits or O(nW) time. Discuss when each is preferable.
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Binary decomposition clicked for me during prep so I could explain it: you split count[i] into powers of two as virtual items, then run standard 0/1 knapsack on the expanded list.
Start by explaining the bounded knapsack problem and why the naive O(nW * count) approach is inefficient. Then describe binary decomposition: split each item's count into powers of two (1, 2, 4, ...) to create new items with weight and value multiplied by the power, reducing the problem to 0/1 knapsack. Finally, analyze the O(nW) time complexity after decomposition and discuss memory trade-offs between 1D and 2D DP arrays.
Pro tip: Emphasize that binary decomposition reduces the number of items to O(log count) per original item, making the solution efficient even for large counts. Also, mention that using a 1D DP array saves memory but requires iterating weights in reverse to avoid reusing items.
Restate the bounded knapsack: each item has a weight, value, and maximum count. Confirm that the goal is to maximize value without exceeding capacity W, and that counts are integers.
For each item with count c, decompose c into powers of two (e.g., 1, 2, 4, ..., remainder). Create new items with weight and value multiplied by these powers, effectively converting the bounded problem into a 0/1 knapsack with O(n log c) items.
Use dynamic programming: dp[j] = max value for capacity j. For each new item, iterate j from W down to weight to avoid reusing items. This yields O(nW) time after decomposition.
Time: O(nW) where n is the number of items after decomposition (original n times log of max count). Space: O(W) with 1D DP, or O(nW) with 2D DP. Discuss trade-offs: 1D saves memory but loses reconstruction ability; 2D allows backtracking but uses more memory.
Mention that if counts are large, binary decomposition is optimal. Also, handle cases where item weight exceeds capacity (skip) and ensure integer overflow is considered for large values.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Honestly the dry-run part was fine for 0/1.
First, clearly define the DP state and recurrence for both 0/1 and bounded knapsack. Then, systematically fill the DP table row by row, showing intermediate values and explaining the decisions at each cell. Finally, extract the optimal value and backtrack to identify the selected items.
Pro tip: Emphasize the difference in state transitions: 0/1 uses a 1D array iterated backwards to avoid reusing items, while bounded knapsack can be optimized with binary splitting or monotonic queue. Mentioning these optimizations shows depth.
State that dp[i][w] represents the maximum value using first i items with capacity w. For 0/1, dp[i][w] = max(dp[i-1][w], dp[i-1][w-weight[i]] + value[i]) if weight[i] <= w. For bounded, include count k: dp[i][w] = max_{0<=k<=count[i], k*weight[i]<=w} (dp[i-1][w - k*weight[i]] + k*value[i]).
Create a table with rows for items 0 to n and columns for capacities 0 to W. Set dp[0][w] = 0 for all w, and dp[i][0] = 0 for all i.
For each item i from 1 to 3, and each capacity w from 0 to 7, compute dp[i][w] using the recurrence. Show the table after each item, highlighting the decision (include or exclude).
Using the same items but with counts [3,1,2], for each item i and capacity w, consider taking k copies (0 <= k <= count[i]) and compute the maximum. Show the updated table.
Read the final answer from dp[n][W]. Then backtrack through the table to determine which items (and how many) were selected, explaining the choices.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.