0 1 Knapsack Problem in DSA C# - 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

↓

Check if item i fits in weight w

↓

Include item i: value = value[i

↓

Exclude item i: value = dp[i-1

↓

Choose max of include/exclude

↓

Update dp table

↓

Repeat for all items and weights

↓

Final answer at dp[last_item

The flow shows how we build a table step-by-step, deciding for each item and weight whether to include the item or not, and storing the best value.

Execution Sample

DSA C#

int[,] dp = new int[4, 8];
int[] weights = {1, 3, 4};
int[] values = {1, 4, 5};
for (int i = 1; i <= 3; i++) {
  for (int w = 1; w <= 7; w++) {
    if (weights[i-1] <= w) {
      dp[i, w] = Math.Max(values[i-1] + dp[i-1, w - weights[i-1]], dp[i-1, w]);
    } else {
      dp[i, w] = dp[i-1, w];
    }
  }
}

This code fills the dp table for 3 items and max weight 7, choosing the best value for each sub-weight.

Execution Table

Step	Item i	Weight w	Include Item Value	Exclude Item Value	Chosen Value	DP Table State (row i)
1.1	1	1	1 + dp[0][0]=1	dp[0][1]=0	1	[0,1,0,0,0,0,0,0]
1.2	1	2	1 + dp[0][1]=1	dp[0][2]=0	1	[0,1,1,0,0,0,0,0]
1.3	1	3	1 + dp[0][2]=1	dp[0][3]=0	1	[0,1,1,1,0,0,0,0]
1.4	1	4	1 + dp[0][3]=1	dp[0][4]=0	1	[0,1,1,1,1,0,0,0]
1.5	1	5	1 + dp[0][4]=1	dp[0][5]=0	1	[0,1,1,1,1,1,0,0]
1.6	1	6	1 + dp[0][5]=1	dp[0][6]=0	1	[0,1,1,1,1,1,1,0]
1.7	1	7	1 + dp[0][6]=1	dp[0][7]=0	1	[0,1,1,1,1,1,1,1]
2.1	2	1	4 + dp[1][-2]=N/A	dp[1][1]=1	1	[0,1,1,1,1,1,1,1]
2.2	2	2	4 + dp[1][-1]=N/A	dp[1][2]=1	1	[0,1,1,1,1,1,1,1]
2.3	2	3	4 + dp[1][0]=4	dp[1][3]=1	4	[0,1,1,4,1,1,1,1]
2.4	2	4	4 + dp[1][1]=5	dp[1][4]=1	5	[0,1,1,4,5,1,1,1]
2.5	2	5	4 + dp[1][2]=5	dp[1][5]=1	5	[0,1,1,4,5,5,1,1]
2.6	2	6	4 + dp[1][3]=5	dp[1][6]=1	5	[0,1,1,4,5,5,5,1]
2.7	2	7	4 + dp[1][4]=5	dp[1][7]=1	5	[0,1,1,4,5,5,5,5]
3.1	3	1	5 + dp[2][-3]=N/A	dp[2][1]=1	1	[0,1,1,4,5,5,5,5]
3.2	3	2	5 + dp[2][-2]=N/A	dp[2][2]=1	1	[0,1,1,4,5,5,5,5]
3.3	3	3	5 + dp[2][-1]=N/A	dp[2][3]=4	4	[0,1,1,4,5,5,5,5]
3.4	3	4	5 + dp[2][0]=5	dp[2][4]=5	5	[0,1,1,4,5,5,5,5]
3.5	3	5	5 + dp[2][1]=6	dp[2][5]=5	6	[0,1,1,4,5,6,5,5]
3.6	3	6	5 + dp[2][2]=6	dp[2][6]=5	6	[0,1,1,4,5,6,6,5]
3.7	3	7	5 + dp[2][3]=9	dp[2][7]=5	9	[0,1,1,4,5,6,6,9]

💡 All items and weights processed; final max value is dp[3][7] = 9

Variable Tracker

Variable	Start	After 1.7	After 2.7	After 3.7
dp[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]	[0,1,1,1,1,1,1,1]
dp[2]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,1,1,4,5,5,5,5]	[0,1,1,4,5,5,5,5]
dp[3]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]	[0,1,1,4,5,6,6,9]

Key Moments - 3 Insights

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

Because if the item weight is more than the current capacity w, we cannot include it. This is shown in execution_table rows like 2.1 and 2.2 where include value is N/A and we choose exclude value.

Why do we add the value of the current item plus dp[i-1][w - weight[i-1]] when including an item?

Because dp[i-1][w - weight[i-1]] represents the best value achievable with previous items and remaining weight after including current item. This is shown in rows like 2.4 where include value is 4 + dp[1][1] = 5.

Why do we take the maximum of including or excluding the item?

Because we want the best total value possible for the current weight. The dp table stores the max value, as seen in the 'Chosen Value' column in execution_table.

Visual Quiz - 3 Questions

Look at the execution table at step 2.4, what is the chosen value for item 2 at weight 4?

A 1

B 4

C 5

D 0

💡 Hint

Check the 'Chosen Value' column in execution_table row 2.4

At which step does the dp table first show a value of 9?

A 3.7

B 2.7

C 1.7

D 3.5

💡 Hint

Look at the 'Chosen Value' and 'DP Table State' columns in execution_table for the highest value

If the weight of item 3 was 5 instead of 4, how would the chosen value at step 3.5 change?

A It would remain 6

B It would become 5

C It would become 9

D It would become 4

💡 Hint

Check how including item 3 depends on weight fitting in current capacity in execution_table rows 3.4 and 3.5

Concept Snapshot

0 1 Knapsack Problem:
- Use a 2D dp table: dp[i][w] = max value using first i items with capacity w
- For each item and weight, decide to include or exclude item
- Include if item weight <= w: dp[i][w] = max(value[i] + dp[i-1][w-weight[i]], dp[i-1][w])
- Else exclude: dp[i][w] = dp[i-1][w]
- Final answer at dp[number_of_items][max_weight]

Full Transcript

The 0 1 Knapsack problem uses a table to find the best value of items that fit in a knapsack of limited weight. We check each item and weight capacity, deciding whether to include the item or not. If the item fits, we compare the value of including it plus the best value for the remaining weight, versus excluding it. We store the best choice in the dp table. After processing all items and weights, the final cell shows the maximum value possible. This process is shown step-by-step in the execution table and variable tracker.

Practice

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

easy

A. The minimum weight of items selected

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

C. The number of items selected regardless of weight

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

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 capacity.

2. Step 2: Clarify constraints

   Each item can be chosen only once (0 or 1 times), and total weight must not exceed the limit.

3. Final Answer:

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

4. Quick Check:

   Maximize value under weight limit = D [OK]

Hint: Remember: maximize value, not weight or count [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Thinking items can be chosen multiple times
• Maximizing weight instead of value

2. Which of the following is the correct C# syntax to declare a 2D array for dynamic programming table in 0 1 Knapsack?

easy

A. int[,] dp = new int[n, W];

B. int[][] dp = new int[n][W];

C. int dp[][] = new int[n][W];

D. int dp = new int[n, W];

Solution

1. Step 1: Recall C# 2D array declaration

   In C#, 2D arrays are declared as int[,] dp = new int[n, W].

2. Step 2: Check each option

   int[,] dp = new int[n, W]; uses correct syntax. int[][] dp = new int[n][W]; is jagged array but syntax is wrong. int dp[][] = new int[n][W]; is invalid in C#. int dp = new int[n, W]; misses brackets.

3. Final Answer:

   int[,] dp = new int[n, W]; -> Option A

4. Quick Check:

   C# 2D array = int[,] dp = new int[n, W]; [OK]

Hint: Use int[,] for 2D arrays in C# [OK]

Common Mistakes:

• Using jagged array syntax incorrectly
• Missing commas in array declaration
• Confusing C# with Java or C++ syntax

3. What is the output of the following C# code snippet for 0 1 Knapsack with weights = {1, 3, 4}, values = {15, 50, 60}, capacity = 5?

int[] weights = {1, 3, 4};
int[] values = {15, 50, 60};
int capacity = 5;
int n = weights.Length;
int[,] dp = new int[n + 1, capacity + 1];
for (int i = 1; i <= n; i++) {
  for (int w = 1; w <= capacity; w++) {
    if (weights[i - 1] <= w) {
      dp[i, w] = Math.Max(values[i - 1] + dp[i - 1, w - weights[i - 1]], dp[i - 1, w]);
    } else {
      dp[i, w] = dp[i - 1, w];
    }
  }
}
Console.WriteLine(dp[n, capacity]);

medium

A. 75

B. 80

C. 65

D. 60

Solution

1. Step 1: Fill dp table for each item and capacity

   For capacity 5 and items: - Item 1 (weight 1, value 15): dp[1,1..5] = 15 - Item 2 (weight 3, value 50): dp[2,3] = max(50 + dp[1,0], dp[1,3]) = 50, dp[2,4] = max(50 + dp[1,1], dp[1,4]) = 65, dp[2,5] = max(50 + dp[1,2], dp[1,5]) = 65 - Item 3 (weight 4, value 60): dp[3,4] = max(60 + dp[2,0], dp[2,4]) = 65, dp[3,5] = max(60 + dp[2,1], dp[2,5]) = 75

2. Step 2: Identify final dp value

   The maximum value for capacity 5 is dp[3,5] = 75.

3. Final Answer:

   75 -> Option A

4. Quick Check:

   Max value at capacity 5 = 75 [OK]

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

Common Mistakes:

• Not adding previous dp values correctly
• Mixing up weights and values
• Off-by-one errors in loops

4. Identify the bug in this C# 0 1 Knapsack code snippet:

int[] weights = {2, 3, 4};
int[] values = {40, 50, 60};
int capacity = 5;
int n = weights.Length;
int[,] dp = new int[n + 1, capacity + 1];
for (int i = 0; i <= n; i++) {
  for (int w = 0; w <= capacity; w++) {
    if (weights[i] <= w) {
      dp[i, w] = Math.Max(values[i] + dp[i - 1, w - weights[i]], dp[i - 1, w]);
    } else {
      dp[i, w] = dp[i - 1, w];
    }
  }
}
Console.WriteLine(dp[n, capacity]);

medium

A. Incorrect comparison operator in if condition

B. Wrong initialization of dp array size

C. Index out of range error due to i starting at 0

D. Missing return statement in method

Solution

1. Step 1: Analyze loop and array indexing

   The loop starts at i=0 and accesses weights[i], values[i], and dp[i-1,...]. At i=0, dp[i-1,...] is dp[-1,...], which is invalid and causes an index error.

2. Step 2: Identify correct loop start

   Loops should start at i=1 to avoid negative indices when accessing dp[i-1,...].

3. Final Answer:

   Index out of range error due to i starting at 0 -> Option C

4. Quick Check:

   Negative index access = error [OK]

Hint: Start loops at 1 when using dp[i-1] indexing [OK]

Common Mistakes:

• Starting loop at 0 causing negative index
• Ignoring dp array bounds
• Confusing loop variables with array indices

5. Given items with weights = {2, 3, 5, 7}, values = {10, 5, 15, 7}, and capacity = 10, which set of items maximizes value without exceeding capacity in 0 1 Knapsack?

hard

A. Items with weights 3 and 7 (total value 12)

B. Items with weights 2, 3, and 5 (total value 30)

C. Items with weights 5 and 7 (total value 22)

D. Items with weights 2 and 7 (total value 17)

Solution

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

   Items with weights 2, 3, and 5 (total value 30): weights 2+3+5=10, values 10+5+15=30 Items with weights 3 and 7 (total value 12): weights 3+7=10, values 5+7=12 Items with weights 5 and 7 (total value 22): weights 5+7=12 (exceeds capacity 10) Items with weights 2 and 7 (total value 17): weights 2+7=9, values 10+7=17

2. Step 2: Choose valid option with max value

   Items with weights 5 and 7 (total value 22) exceeds capacity, so invalid. Among others, Items with weights 2, 3, and 5 (total value 30) has max value 30.

3. Final Answer:

   Items with weights 2, 3, and 5 (total value 30) -> Option B

4. Quick Check:

   Max value under capacity 10 = 30 [OK]

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

Common Mistakes:

• Choosing items exceeding capacity
• Ignoring total value calculation
• Not checking all combinations