Start by defining the DP state and transition for the standard 0/1 knapsack, then explain how to compress the 2D DP to 1D by iterating weights in reverse. For reconstruction, describe how to store parent pointers or use a divide-and-conquer technique to recover the chosen items without breaking the O(W) space constraint.
Pro tip: Mention that while O(W) space is possible, storing parent pointers for each state may increase space to O(nW); instead, use a recursive divide-and-conquer approach or store only the decisions at each step to achieve O(W) space with O(nW) time.
Explain that dp[j] represents the maximum value achievable with capacity j using the first i items. The transition is dp[j] = max(dp[j], dp[j - weight[i]] + value[i]) for j from W down to weight[i].
Show how to compute the maximum value in O(nW) time and O(W) space by iterating items and updating the dp array in reverse order to avoid reusing items.
Discuss that the 1D DP only stores the optimal value, not the choices. To reconstruct, we need additional information or a different strategy.
Present a divide-and-conquer approach: split items into two halves, compute DP for each half, then find the split point that yields the optimal value and recursively reconstruct each half. Alternatively, store parent pointers in a separate array but note the space trade-off.
Confirm that the divide-and-conquer method uses O(W) space and O(nW) time, and discuss the trade-off between space and simplicity when using parent pointers.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Unbounded I was fine with, that's the easier mental shift since you just don't exclude the current item from consideration after picking it.
Start by briefly restating the classic 0/1 knapsack DP and its complexity, then explain how the recurrence and iteration order change for unbounded (allow multiple items) and bounded (limited copies) variants. Conclude by comparing time and space complexities, noting optimizations like 1D arrays and binary splitting.
Pro tip: Emphasize that the unbounded variant uses forward iteration in the 1D DP to allow reuse, while 0/1 uses backward iteration to avoid reuse—this subtlety often trips up candidates. Also mention that bounded knapsack can be optimized with binary splitting to achieve O(nW log C) time, which is a common follow-up.
State the standard DP: dp[i][w] = max(dp[i-1][w], dp[i-1][w-wt[i]] + val[i]), with O(nW) time and space. Mention the 1D optimization using backward iteration.
Change the recurrence to allow multiple uses: dp[w] = max(dp[w], dp[w-wt[i]] + val[i]) and iterate w forward. Time remains O(nW), space O(W).
Introduce a count limit per item. Use a 2D DP with an extra dimension for count, or optimize with binary splitting into 0/1 items, reducing time to O(nW log C) and space O(W).
Summarize: 0/1 and unbounded are O(nW) time, O(W) space (with 1D). Bounded is O(nW * maxCount) naively, or O(nW log C) with binary splitting. Space remains O(W) for all with optimization.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.