0 1 Knapsack Problem in DSA Typescript - Execution Trace

Concept Flow - 0 1 Knapsack Problem

Start with empty DP table

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For each item i

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For each weight w from 0 to max

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Check if item i fits in weight w

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Include item i: value = val[i

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Exclude item i: value = dp[i-1

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Choose max(include, exclude)

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Fill dp[i

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Repeat for all items and weights

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Answer = dp[last_item

We build a table where each cell shows the best value for items up to i and weight w, choosing to include or exclude the current item.

Execution Sample

DSA Typescript

weights = [1, 3, 4]
values = [1, 4, 5]
maxWeight = 7
// dp table rows: items, cols: weights
// Fill dp step by step

This code fills a table to find max value for knapsack capacity 7 with 3 items.

Execution Table

Step	Item	Weight Capacity w	Include Item Value	Exclude Item Value	Chosen Value	DP Table Row State
1	Item 1 (w=1,v=1)	w=0	N/A (item too heavy)	dp[0][0]=0	0	[0, 0, 0, 0, 0, 0, 0, 0]
2	Item 1 (w=1,v=1)	w=1	1 + dp[0][0]=1	dp[0][1]=0	1	[0, 1, 0, 0, 0, 0, 0, 0]
3	Item 1 (w=1,v=1)	w=2	1 + dp[0][1]=1	dp[0][2]=0	1	[0, 1, 1, 0, 0, 0, 0, 0]
4	Item 1 (w=1,v=1)	w=3	1 + dp[0][2]=1	dp[0][3]=0	1	[0, 1, 1, 1, 0, 0, 0, 0]
5	Item 1 (w=1,v=1)	w=4	1 + dp[0][3]=1	dp[0][4]=0	1	[0, 1, 1, 1, 1, 0, 0, 0]
6	Item 1 (w=1,v=1)	w=5	1 + dp[0][4]=1	dp[0][5]=0	1	[0, 1, 1, 1, 1, 1, 0, 0]
7	Item 1 (w=1,v=1)	w=6	1 + dp[0][5]=1	dp[0][6]=0	1	[0, 1, 1, 1, 1, 1, 1, 0]
8	Item 1 (w=1,v=1)	w=7	1 + dp[0][6]=1	dp[0][7]=0	1	[0, 1, 1, 1, 1, 1, 1, 1]
9	Item 2 (w=3,v=4)	w=0	N/A	dp[1][0]=0	0	[0, 1, 1, 1, 1, 1, 1, 1]
10	Item 2 (w=3,v=4)	w=1	N/A	dp[1][1]=1	1	[0, 1, 1, 1, 1, 1, 1, 1]
11	Item 2 (w=3,v=4)	w=2	N/A	dp[1][2]=1	1	[0, 1, 1, 1, 1, 1, 1, 1]
12	Item 2 (w=3,v=4)	w=3	4 + dp[0][0]=4	dp[0][3]=1	4	[0, 1, 1, 4, 1, 1, 1, 1]
13	Item 2 (w=3,v=4)	w=4	4 + dp[0][1]=5	dp[0][4]=1	5	[0, 1, 1, 4, 5, 1, 1, 1]
14	Item 2 (w=3,v=4)	w=5	4 + dp[0][2]=5	dp[0][5]=1	5	[0, 1, 1, 4, 5, 5, 1, 1]
15	Item 2 (w=3,v=4)	w=6	4 + dp[0][3]=5	dp[0][6]=1	5	[0, 1, 1, 4, 5, 5, 5, 1]
16	Item 2 (w=3,v=4)	w=7	4 + dp[0][4]=6	dp[0][7]=1	6	[0, 1, 1, 4, 5, 5, 5, 6]
17	Item 3 (w=4,v=5)	w=0	N/A	dp[2][0]=0	0	[0, 1, 1, 4, 5, 5, 5, 6]
18	Item 3 (w=4,v=5)	w=1	N/A	dp[2][1]=1	1	[0, 1, 1, 4, 5, 5, 5, 6]
19	Item 3 (w=4,v=5)	w=2	N/A	dp[2][2]=1	1	[0, 1, 1, 4, 5, 5, 5, 6]
20	Item 3 (w=4,v=5)	w=3	N/A	dp[2][3]=4	4	[0, 1, 1, 4, 5, 5, 5, 6]
21	Item 3 (w=4,v=5)	w=4	5 + dp[1][0]=5	dp[1][4]=5	5	[0, 1, 1, 4, 5, 5, 5, 6]
22	Item 3 (w=4,v=5)	w=5	5 + dp[1][1]=6	dp[1][5]=5	6	[0, 1, 1, 4, 5, 6, 5, 6]
23	Item 3 (w=4,v=5)	w=6	5 + dp[1][2]=6	dp[1][6]=5	6	[0, 1, 1, 4, 5, 6, 6, 6]
24	Item 3 (w=4,v=5)	w=7	5 + dp[1][3]=9	dp[1][7]=6	9	[0, 1, 1, 4, 5, 6, 6, 9]

💡 All items and weights processed; final answer dp[3][7] = 9

Variable Tracker

Variable	Start	After Item 1	After Item 2	After Item 3	Final
dp Table Row	[0,0,0,0,0,0,0,0]	[0,1,1,1,1,1,1,1]	[0,1,1,4,5,5,5,6]	[0,1,1,4,5,6,6,9]	[0,1,1,4,5,6,6,9]
Current Item	None	Item 1	Item 2	Item 3	Done
Weight Capacity w	0	0-7	0-7	0-7	7

Key Moments - 3 Insights

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

Because dp[i-1][w - weight[i]] holds the best value without the current item for the remaining weight, so adding current item's value gives total value if included (see steps 12, 13, 22).

Why do we sometimes have 'N/A' for include item value?

When the item's weight is more than current capacity w, it can't fit, so we can't include it (see steps 1, 9, 17).

Why do we choose the max between including and excluding the item?

Because we want the best total value possible for that weight and items considered (see column 'Chosen Value' in execution_table).

Visual Quiz - 3 Questions

Look at the execution table at step 24, what is the chosen value for weight 7 after including item 3?

A 9

B 6

C 5

D 4

💡 Hint

Check the 'Chosen Value' column at step 24 for weight 7.

At which step does the condition 'item fits in weight' become false for item 2?

A Step 9

B Step 12

C Step 10

D Step 11

💡 Hint

Look at steps for item 2 where include value is 'N/A'.

If the weight of item 3 was 5 instead of 4, how would the chosen value at weight 7 change at step 24?

A It would remain 9

B It would decrease

C It would increase

D It would be zero

💡 Hint

Consider if item 3 fits in weight 7 when weight is 5, and check how include value changes.

Concept Snapshot

0 1 Knapsack Problem:
- Use a 2D dp table: rows = items, cols = weights
- dp[i][w] = max value using first i items with capacity w
- For each item and weight:
  Include item if fits: val[i] + dp[i-1][w - weight[i]]
  Exclude item: dp[i-1][w]
- Choose max of include/exclude
- Final answer at dp[last_item][maxWeight]

Full Transcript

The 0 1 Knapsack problem uses a table to find the best value for a given weight limit and items. We fill the table row by row for each item and column by column for each weight capacity. For each cell, we check if the current item fits. If yes, we consider including it by adding its value plus the best value for the remaining weight. We also consider excluding it by taking the previous best value without this item. We pick the maximum of these two. This process repeats until all items and weights are processed. The final answer is in the last cell of the table. This method ensures we consider all combinations without repetition.