0 1 Knapsack Problem in DSA C - Execution Trace

Concept Flow - 0 1 Knapsack Problem

Start with empty knapsack

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For each item i

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For each weight w from 0 to W

↓

Check if item i fits (weight[i

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Include item i: value[i

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Update dp[i

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Repeat for all items and weights

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Final answer in dp[n

We fill a table dp where each cell dp[i][w] stores the best value using first i items with max weight w. For each item and weight, we decide to include or exclude the item to maximize value.

Execution Sample

DSA C

int dp[4][8] = {0};
int weights[] = {1, 3, 4};
int values[] = {1, 4, 5};
int W = 7;

// Fill dp table

This code fills a dp table for 3 items and max weight 7 to find max value.

Execution Table

Step	Operation	Item i	Weight w	Include Item Value	Exclude Item Value	dp[i][w] Updated	dp Table State
1	Initialize dp row 0	0	0-7	-	-	0	[0,0,0,0,0,0,0,0]
2	Process item 1	1	0	-	dp[0][0]=0	0	[0,0,0,0,0,0,0,0]
3	Process item 1	1	1	value=1 + dp[0][0]=0	dp[0][1]=0	1	[0,1,0,0,0,0,0,0]
4	Process item 1	1	2	value=1 + dp[0][1]=0	dp[0][2]=0	1	[0,1,1,0,0,0,0,0]
5	Process item 1	1	3	value=1 + dp[0][2]=0	dp[0][3]=0	1	[0,1,1,1,0,0,0,0]
6	Process item 1	1	4	value=1 + dp[0][3]=0	dp[0][4]=0	1	[0,1,1,1,1,0,0,0]
7	Process item 1	1	5	value=1 + dp[0][4]=0	dp[0][5]=0	1	[0,1,1,1,1,1,0,0]
8	Process item 1	1	6	value=1 + dp[0][5]=0	dp[0][6]=0	1	[0,1,1,1,1,1,1,0]
9	Process item 1	1	7	value=1 + dp[0][6]=0	dp[0][7]=0	1	[0,1,1,1,1,1,1,1]
10	Process item 2	2	0	-	dp[1][0]=0	0	[0,1,1,1,1,1,1,1]
11	Process item 2	2	1	-	dp[1][1]=1	1	[0,1,1,1,1,1,1,1]
12	Process item 2	2	2	-	dp[1][2]=1	1	[0,1,1,1,1,1,1,1]
13	Process item 2	2	3	value=4 + dp[1][0]=0	dp[1][3]=1	4	[0,1,1,4,1,1,1,1]
14	Process item 2	2	4	value=4 + dp[1][1]=1	dp[1][4]=1	5	[0,1,1,4,5,1,1,1]
15	Process item 2	2	5	value=4 + dp[1][2]=1	dp[1][5]=1	5	[0,1,1,4,5,5,1,1]
16	Process item 2	2	6	value=4 + dp[1][3]=4	dp[1][6]=1	8	[0,1,1,4,5,5,8,1]
17	Process item 2	2	7	value=4 + dp[1][4]=5	dp[1][7]=1	9	[0,1,1,4,5,5,8,9]
18	Process item 3	3	0	-	dp[2][0]=0	0	[0,1,1,4,5,5,8,9]
19	Process item 3	3	1	-	dp[2][1]=1	1	[0,1,1,4,5,5,8,9]
20	Process item 3	3	2	-	dp[2][2]=1	1	[0,1,1,4,5,5,8,9]
21	Process item 3	3	3	-	dp[2][3]=4	4	[0,1,1,4,5,5,8,9]
22	Process item 3	3	4	value=5 + dp[2][0]=0	dp[2][4]=5	5	[0,1,1,4,5,5,8,9]
23	Process item 3	3	5	value=5 + dp[2][1]=1	dp[2][5]=5	6	[0,1,1,4,5,6,8,9]
24	Process item 3	3	6	value=5 + dp[2][2]=1	dp[2][6]=8	6	[0,1,1,4,5,6,8,9]
25	Process item 3	3	7	value=5 + dp[2][3]=4	dp[2][7]=9	9	[0,1,1,4,5,6,8,9]
26	End	-	-	-	-	-	Final dp table filled

💡 All items processed for all weights up to W=7, dp table complete

Variable Tracker

Variable	Start	After Step 9	After Step 17	After Step 25	Final
dp row	[0,0,0,0,0,0,0,0]	[0,1,1,1,1,1,1,1]	[0,1,1,4,5,5,8,9]	[0,1,1,4,5,6,8,9]	[0,1,1,4,5,6,8,9]
i (item index)	0	1	2	3	3
w (weight)	0	7	7	7	7

Key Moments - 3 Insights

Why do we add value[i] to dp[i-1][w-weight[i]] when including an item?

Because dp[i-1][w-weight[i]] represents the best value achievable with previous items and remaining weight after including current item. See rows 13,14 where inclusion sums value[i] + dp from smaller weight.

Why do we sometimes keep dp[i][w] same as dp[i-1][w]?

When the item doesn't fit or excluding it gives better value, we keep dp[i][w] same as dp[i-1][w]. For example, in rows 2,3, and 10,11 exclusion is chosen.

Why does the dp table have one extra row and column?

Row 0 and column 0 represent base cases: no items or zero capacity. This simplifies logic by providing initial values, as seen in step 1.

Visual Quiz - 3 Questions

Look at the execution_table at step 16, what is the value of dp[2][6]?

A 6

B 5

C 8

D 9

💡 Hint

Check the 'dp[i][w] Updated' column at step 16 in execution_table

At which step does the dp table first update dp[i][w] to 4?

A Step 21

B Step 13

C Step 14

D Step 22

💡 Hint

Look for dp[i][w] Updated column with value 4 in execution_table

If the weight of item 2 changed from 3 to 2, how would dp[2][3] change at step 13?

A It would become 5

B It would remain 4

C It would become 1

D It would become 0

💡 Hint

Including item 2 at weight 3 would add value[2] + dp[i-1][w-weight[i]]; with smaller weight, dp index changes

Concept Snapshot

0 1 Knapsack Problem:
- Use dp[i][w] for max value with first i items and weight limit w
- For each item and weight, choose max(include, exclude)
- Include: value[i] + dp[i-1][w-weight[i]] if fits
- Exclude: dp[i-1][w]
- Final answer at dp[n][W]

Full Transcript

The 0 1 Knapsack problem uses a table dp where dp[i][w] stores the best value using first i items with max weight w. We fill this table row by row, checking for each item if it fits in the current weight. If it fits, we consider including it by adding its value plus the best value for remaining weight. Otherwise, we exclude it and keep previous best. The final answer is in dp[n][W]. The execution table shows step-by-step how dp is updated for each item and weight. Key moments clarify why we add values and why base cases exist. The visual quiz tests understanding of dp values at specific steps and effects of weight changes.

Practice

1. In the 0/1 Knapsack problem, why can't we take fractions of items like in the Fractional Knapsack?

easy

A. Because the DP table only stores integers

B. Because each item must be taken as a whole unit or not at all - fractional inclusion violates the problem definition

C. Because fractions make the recursion infinite

D. Because the weight array only accepts integer values

Solution

1. 0/1 Constraint

   The '0/1' in the name literally means each item has two states: not taken (0) or fully taken (1). There is no state for 0.5 or any fraction.

2. Why Greedy Fails

   Because you cannot take fractions, the greedy 'highest value/weight ratio first' strategy fails - taking a high-ratio item may block a better combination of whole items.

3. Why DP is Needed

   DP explores all combinations of 0/1 choices, which is required when greedy fails due to the binary selection constraint.

4. Final Answer:

   Each item is binary - take it whole or skip it. → Option B

Hint: 0/1 = binary choice per item: take it whole or skip entirely. [OK]

Common Mistakes:

• Confusing 0/1 Knapsack with Fractional Knapsack
• Thinking DP table data type restricts selection
• Assuming greedy works for 0/1 Knapsack

2. For 0/1 Knapsack with n=3 items and capacity W=5, what is the size of the standard DP table dp[n+1][W+1] in C?

easy

A. 3 × 5

B. 4 × 5

C. 3 × 6

D. 4 × 6

Solution

1. Table Dimensions

   The DP table is dp[n+1][W+1]. With n=3 items and W=5 capacity: dp[4][6].

2. Why +1

   Index 0 represents the base case: dp[0][w] = 0 (no items) and dp[i][0] = 0 (zero capacity). The +1 makes these base cases fit naturally without off-by-one adjustments.

3. In C

   int dp[4][6]; - 4 rows (0 to 3 items), 6 columns (capacity 0 to 5).

4. Final Answer:

   4 × 6 → Option D

Hint: dp[n+1][W+1] → always add 1 to both dimensions for the base case. [OK]

Common Mistakes:

• Using dp[n][W] and missing the base case row/column
• Adding +1 to only one dimension
• Confusing n items with n+1 rows

3. In the 0/1 Knapsack DP recurrence, when weight[i] > w (current item is heavier than current capacity), what is dp[i][w]?

medium

A. 0

B. dp[i-1][w] - same value as without this item

C. dp[i][w-1]

D. value[i] + dp[i-1][w - weight[i]]

Solution

1. Can't Include Overweight Item

   If weight[i] > w, item i cannot fit - we must skip it.

2. Inherit from Previous Row

   dp[i][w] = dp[i-1][w]: the best value with first i items equals the best with first i-1 items.

3. Full Recurrence

   if (weight[i] > w): dp[i][w] = dp[i-1][w]. else: dp[i][w] = max(dp[i-1][w], value[i] + dp[i-1][w-weight[i]])

4. Final Answer:

   dp[i-1][w] → Option B

Hint: Item too heavy → skip → copy from row above: dp[i-1][w]. [OK]

Common Mistakes:

• Setting dp[i][w]=0 instead of carrying forward the previous best
• Applying include formula when item doesn't fit
• Reading dp[i][w-1] instead of dp[i-1][w]

4. What is the time and space complexity of the standard bottom-up 0/1 Knapsack DP solution?

medium

A. Time O(n×W), Space O(n×W)

B. Time O(n×W), Space O(W)

C. Time O(2^n), Space O(n)

D. Time O(n log W), Space O(n×W)

Solution

1. Time Complexity

   Two nested loops: outer over n items, inner over W+1 capacities. Each cell O(1). Total: O(n×W).

2. Space Complexity

   Full table dp[n+1][W+1] → O(n×W) space.

3. Note

   Space can be reduced to O(W) with 1D rolling array, but standard 2D table is O(n×W).

4. Final Answer:

   Time O(n×W), Space O(n×W) → Option A

Hint: Standard 2D table = O(n×W) time AND space. 1D optimized = O(W) space only. [OK]

Common Mistakes:

• Confusing standard O(n×W) space with optimized O(W)
• Claiming O(2^n) which applies only to naive recursion
• Mixing up time and space complexities

5. In the space-optimized 1D 0/1 Knapsack in C, why must the inner capacity loop iterate from W down to weight[i], not from 0 up to W?

hard

A. To avoid negative array indices

B. To prevent using item i more than once - forward iteration would allow the same item to be counted multiple times in the same pass

C. Because C arrays are stored in column-major order

D. To match the order of the weight array

Solution

1. Problem With Left-to-Right

   With a 1D array, processing left to right means dp[w-weight[i]] has already been updated in this pass - item i gets counted again at smaller capacities.

2. Right-to-Left Prevents Reuse

   Iterating from W down to weight[i] ensures dp[w-weight[i]] still holds the pre-item-i value. Each item counted at most once.

3. 0/1 vs Unbounded

   Right-to-left = 0/1 Knapsack. Left-to-right = Unbounded Knapsack.

4. Final Answer:

   Prevents reusing item i in the same pass. → Option B

Hint: Right-to-left = 0/1 (no reuse). Left-to-right = Unbounded (reuse allowed). [OK]

Common Mistakes:

• Using left-to-right which implements Unbounded Knapsack accidentally
• Thinking loop direction is about array bounds not item reuse
• Not knowing right-to-left is the key 0/1 vs Unbounded distinction