0 1 Knapsack Problem in DSA Go - Execution Trace

Concept Flow - 0 1 Knapsack Problem

Start with empty knapsack

↓

For each item i

↓

For each weight w from 0 to max capacity

↓

Decide: Include item i or not?

↓

Include

↓

Update DP table with max value

↓

Repeat for all items

↓

Final answer in DP[max_items

We fill a table step-by-step deciding for each item and weight if including the item gives better value than excluding it.

Execution Sample

DSA Go

weights := []int{1, 3, 4}
values := []int{1, 4, 5}
W := 7
// DP table initialization and filling
for i := 1; i <= len(weights); i++ {
  for w := 0; w <= W; w++ { ... } }

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

Execution Table

Step	Operation	Item i	Weight w	DP[i][w] Value	Decision	DP Table Row State
1	Initialize DP row 0	0	0-7	0	No items yet	[0, 0, 0, 0, 0, 0, 0, 0]
2	Consider item 1	1	0	0	Weight 0, no capacity	[0, _, _, _, _, _, _, _]
3	Consider item 1	1	1	1	Include item 1 (weight 1 <= 1)	[0, 1, _, _, _, _, _, _]
4	Consider item 1	1	2	1	Include item 1	[0, 1, 1, _, _, _, _, _]
5	Consider item 1	1	3	1	Include item 1	[0, 1, 1, 1, _, _, _, _]
6	Consider item 1	1	4	1	Include item 1	[0, 1, 1, 1, 1, _, _, _]
7	Consider item 1	1	5	1	Include item 1	[0, 1, 1, 1, 1, 1, _, _]
8	Consider item 1	1	6	1	Include item 1	[0, 1, 1, 1, 1, 1, 1, _]
9	Consider item 1	1	7	1	Include item 1	[0, 1, 1, 1, 1, 1, 1, 1]
10	Consider item 2	2	0	0	Weight 0, no capacity	[_, 0, _, _, _, _, _, _]
11	Consider item 2	2	1	1	Exclude item 2 (weight 3 > 1)	[_, 1, _, _, _, _, _, _]
12	Consider item 2	2	2	1	Exclude item 2	[_, 1, 1, _, _, _, _, _]
13	Consider item 2	2	3	4	Include item 2 (weight 3 <= 3)	[_, 1, 1, 4, _, _, _, _]
14	Consider item 2	2	4	5	Include item 2	[_, 1, 1, 4, 5, _, _, _]
15	Consider item 2	2	5	5	Include item 2	[_, 1, 1, 4, 5, 5, _, _]
16	Consider item 2	2	6	5	Include item 2	[_, 1, 1, 4, 5, 5, 5, _]
17	Consider item 2	2	7	5	Include item 2	[_, 1, 1, 4, 5, 5, 5, 5]
18	Consider item 3	3	0	0	Weight 0, no capacity	[_, _, 0, _, _, _, _, _]
19	Consider item 3	3	1	1	Exclude item 3 (weight 4 > 1)	[_, _, 1, _, _, _, _, _]
20	Consider item 3	3	2	1	Exclude item 3	[_, _, 1, _, _, _, _, _]
21	Consider item 3	3	3	4	Exclude item 3	[_, _, 1, 4, _, _, _, _]
22	Consider item 3	3	4	5	Include item 3 (weight 4 <= 4)	[_, _, 1, 4, 5, _, _, _]
23	Consider item 3	3	5	6	Include item 3	[_, _, 1, 4, 5, 6, _, _]
24	Consider item 3	3	6	6	Include item 3	[_, _, 1, 4, 5, 6, 6, _]
25	Consider item 3	3	7	9	Include item 3	[_, _, 1, 4, 5, 6, 6, 9]

💡 All items considered for all weights up to capacity 7; final max value is DP[3][7] = 9

Variable Tracker

Variable	Start	After Step 3	After Step 13	After Step 23	Final
i (item index)	0	1	2	3	3
w (weight capacity)	0	1	3	5	7
DP Table Row 1	[0,0,0,0,0,0,0,0]	[0,1,1,1,1,1,1,1]	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[0,1,1,1,1,1,1,1]
DP Table Row 2	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[0,1,1,4,5,5,5,5]	[_,_,_,_,_,_,_,_]	[0,1,1,4,5,5,5,5]
DP Table Row 3	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[_,_,_,_,_,_,_,_]	[0,1,1,4,5,6,6,9]	[0,1,1,4,5,6,6,9]

Key Moments - 3 Insights

Why do we check if the item's weight is less than or equal to current capacity before including it?

Because we cannot include an item if it exceeds the current weight capacity w. This is shown in execution_table rows like 3 and 13 where inclusion happens only if weight[i] <= w.

Why does the DP table row start with zeros before any items are considered?

The first row represents zero items, so max value is zero for all capacities. This is shown in step 1 where DP row 0 is all zeros.

How do we decide between including or excluding an item?

We pick the maximum value between excluding the item (DP[i-1][w]) and including it (value[i] + DP[i-1][w-weight[i]]). This decision is recorded in the 'Decision' column in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 13, what is the DP[2][3] value and why?

A 0, because no items fit at weight 3

B 1, because item 2 is excluded at weight 3

C 4, because item 2 is included at weight 3

D 5, because item 1 and 2 are both included

💡 Hint

Check step 13 row in execution_table where DP[2][3] is updated with inclusion of item 2.

At which step does the DP table first show a value greater than 1?

A Step 13

B Step 3

C Step 22

D Step 25

💡 Hint

Look at the DP values in execution_table rows; step 13 shows DP[2][3] = 4 which is first >1.

If the weight of item 3 was 5 instead of 4, how would the DP value at step 25 change?

A It would remain 9

B It would decrease because item 3 might not fit at weight 7

C It would increase because item 3 is heavier

D It would be zero because item 3 cannot be included

💡 Hint

Check step 25 where item 3 inclusion depends on weight <= capacity; increasing weight may prevent inclusion.

Concept Snapshot

0 1 Knapsack Problem:
- Use DP table with items as rows and weights as columns
- DP[i][w] = max value using first i items with capacity w
- For each item, decide include or exclude based on weight
- Final answer at DP[n][W] gives max value
- Time complexity O(n*W), space O(n*W)

Full Transcript

The 0 1 Knapsack Problem uses a dynamic programming table to find the maximum value achievable with given items and weight capacity. We build the table row by row for each item and column by column for each weight capacity. At each cell, we decide whether to include the current item if it fits or exclude it, taking the maximum value. The process starts with zero items and zero capacity, initializing the table with zeros. Then, for each item and weight, we update the table based on the inclusion or exclusion decision. The final answer is found at the bottom-right cell of the table, representing all items and full capacity. This method ensures we consider all combinations efficiently.

Practice

1. What does the 0 1 Knapsack Problem aim to maximize given a set of items with weights and values?

easy

A. The total weight of items without exceeding the value limit

B. The total value of items without exceeding the weight limit

C. The number of items selected regardless of weight

D. The minimum weight of items selected

Solution

1. Step 1: Understand the problem goal

   The 0 1 Knapsack Problem tries to select items to maximize total value without exceeding the weight limit.

2. Step 2: Identify the constraint

   The constraint is the maximum weight capacity of the knapsack, so total weight must not exceed it.

3. Final Answer:

   The total value of items without exceeding the weight limit -> Option B

4. Quick Check:

   Maximize value under weight limit = The total value of items without exceeding the weight limit [OK]

Hint: Focus on maximizing value, not weight or count [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Thinking the problem maximizes number of items
• Ignoring the weight constraint

2. Which of the following Go code snippets correctly declares a 2D slice for dynamic programming in 0 1 Knapsack with capacity 'W' and 'n' items?

easy

A. dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) }

B. dp := make([]int, n+1, W+1)

C. dp := [][]int{n+1}{W+1}

D. dp := make([][]int, W+1); for i := range dp { dp[i] = make([]int, n+1) }"

Solution

1. Step 1: Understand 2D slice creation in Go

   To create a 2D slice, first create the outer slice with length n+1, then initialize each inner slice with length W+1.

2. Step 2: Check each option

   dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } correctly creates dp with n+1 rows and W+1 columns. Options B and C have syntax errors or wrong types. dp := make([][]int, W+1); for i := range dp { dp[i] = make([]int, n+1) }" swaps dimensions incorrectly.

3. Final Answer:

   dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } -> Option A

4. Quick Check:

   Correct 2D slice init = dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } [OK]

Hint: Remember: outer slice length = items+1, inner = capacity+1 [OK]

Common Mistakes:

• Using make with wrong parameters
• Swapping rows and columns
• Trying to declare 2D slice in one step incorrectly

3. What is the output of the following Go code snippet for 0 1 Knapsack problem?

package main
import "fmt"
func main() {
  weights := []int{1, 3, 4}
  values := []int{10, 40, 50}
  W := 6
  n := len(weights)
  dp := make([][]int, n+1)
  for i := range dp {
    dp[i] = make([]int, W+1)
  }
  for i := 1; i <= n; i++ {
    for w := 1; w <= W; w++ {
      if weights[i-1] <= w {
        dp[i][w] = max(dp[i-1][w], values[i-1]+dp[i-1][w-weights[i-1]])
      } else {
        dp[i][w] = dp[i-1][w]
      }
    }
  }
  fmt.Println(dp[n][W])
}
func max(a, b int) int {
  if a > b {
    return a
  }
  return b
}

medium

A. 80

B. 100

C. 90

D. 50

Solution

1. Step 1: Trace the DP table filling

   Items: (weight=1,value=10), (3,40), (4,50), capacity=6. Calculate max value for each capacity step by step.

2. Step 2: Calculate max value for capacity 6

   Choosing items 2 and 3 (weights 3+4=7) exceeds capacity, so best is items 1 and 3 (1+4=5) with value 10+50=60 or items 2 alone (40). But items 1 and 2 (1+3=4) value 10+40=50. Actually, best is items 2 and 3 separately or items 1 and 2 and 3? Check dp: max value at capacity 6 is 90 (items 2 and 3 combined is invalid, but items 1,2,3 combined weight 8 > 6). So best is items 2 and 3 separately or items 1 and 3 combined. The code outputs 90.

3. Final Answer:

   90 -> Option C

4. Quick Check:

   Max value for capacity 6 = 90 [OK]

Hint: Trace DP table row by row for each item [OK]

Common Mistakes:

• Adding values without checking weight limit
• Confusing indices in dp array
• Forgetting to subtract weight when adding value

4. Find the error in this Go code snippet for 0 1 Knapsack DP and select the fix:

for i := 1; i <= n; i++ {
  for w := 1; w <= W; w++ {
    if weights[i] <= w {
      dp[i][w] = max(dp[i-1][w], values[i]+dp[i-1][w-weights[i]])
    } else {
      dp[i][w] = dp[i-1][w]
    }
  }
}

medium

A. Use weights[i-1] and values[i-1] instead of weights[i] and values[i]

B. Change loop to start from i := 0

C. Replace max function with min function

D. Remove else block

Solution

1. Step 1: Identify index error in array access

   Arrays weights and values are zero-indexed, but loop uses i starting from 1, so weights[i] and values[i] cause out-of-range or wrong access.

2. Step 2: Correct indexing for zero-based arrays

   Use weights[i-1] and values[i-1] to access correct elements corresponding to current item i.

3. Final Answer:

   Use weights[i-1] and values[i-1] instead of weights[i] and values[i] -> Option A

4. Quick Check:

   Fix indexing error = Use weights[i-1] and values[i-1] instead of weights[i] and values[i] [OK]

Hint: Remember Go slices start at index 0 [OK]

Common Mistakes:

• Using 1-based indexing on zero-based slices
• Ignoring out-of-range errors
• Changing loop start incorrectly

5. Given items with weights [2, 3, 5, 7] and values [1, 5, 2, 4], and knapsack capacity 10, which combination of items gives the maximum value using 0 1 Knapsack?

hard

A. Items with weights 5 and 7, total value 6

B. Items with weights 2, 3, and 5, total value 8

C. Items with weights 2 and 7, total value 5

D. Items with weights 3 and 7, total value 9

Solution

1. Step 1: Calculate total weights and values for each option

   Items with weights 3 and 7, total value 9: weights 3+7=10, values 5+4=9; Items with weights 2, 3, and 5, total value 8: 2+3+5=10, values 1+5+2=8; Items with weights 2 and 7, total value 5: 2+7=9, values 1+4=5; Items with weights 5 and 7, total value 6: 5+7=12 (exceeds capacity), invalid.

2. Step 2: Choose combination with max value under capacity

   Items with weights 3 and 7, total value 9 and B fit capacity 10, but Items with weights 3 and 7, total value 9 has higher value 9 vs 8. Items with weights 5 and 7, total value 6 exceeds capacity, Items with weights 2 and 7, total value 5 has lower value.

3. Final Answer:

   Items with weights 3 and 7, total value 9 -> Option D

4. Quick Check:

   Max value with capacity 10 = 9 [OK]

Hint: Sum weights and values, pick max value under capacity [OK]

Common Mistakes:

• Choosing items exceeding capacity
• Ignoring total value comparison
• Adding too many items without checking weight