Unbounded Knapsack Problem in DSA C# - Execution Trace

Concept Flow - Unbounded Knapsack Problem

Start

↓

Initialize DP array with 0

↓

For each weight from 1 to W

↓

For each item

↓

If item weight <= current weight

↓

Update DP[current weight

↓

Repeat for all weights

↓

DP[W

↓

End

We fill a DP array from weight 0 to W, at each weight checking all items to find the best value by possibly including items multiple times.

Execution Sample

DSA C#

int[] dp = new int[W + 1];
for (int w = 1; w <= W; w++) {
  for (int i = 0; i < items.Length; i++) {
    if (items[i].weight <= w) {
      dp[w] = Math.Max(dp[w], dp[w - items[i].weight] + items[i].value);
    }
  }
}

This code fills the dp array to find the max value for each weight up to W using unbounded items.

Execution Table

Step	Operation	Current Weight (w)	Item Considered (weight,value)	DP Update	DP Array State
1	Initialize dp array	-	-	dp = [0,0,0,0,0,0,0,0]	[0,0,0,0,0,0,0,0]
2	w=1, i=0	1	(1,1)	dp[1] = max(dp[1], dp[1-1]+1) = max(0,0+1)=1	[0,1,0,0,0,0,0,0]
3	w=1, i=1	1	(3,4)	3 > 1 skip	[0,1,0,0,0,0,0,0]
4	w=2, i=0	2	(1,1)	dp[2] = max(dp[2], dp[2-1]+1) = max(0,1+1)=2	[0,1,2,0,0,0,0,0]
5	w=2, i=1	2	(3,4)	3 > 2 skip	[0,1,2,0,0,0,0,0]
6	w=3, i=0	3	(1,1)	dp[3] = max(dp[3], dp[3-1]+1) = max(0,2+1)=3	[0,1,2,3,0,0,0,0]
7	w=3, i=1	3	(3,4)	dp[3] = max(3, dp[3-3]+4) = max(3,0+4)=4	[0,1,2,4,0,0,0,0]
8	w=4, i=0	4	(1,1)	dp[4] = max(dp[4], dp[4-1]+1) = max(0,4+1)=5	[0,1,2,4,5,0,0,0]
9	w=4, i=1	4	(3,4)	dp[4] = max(5, dp[4-3]+4) = max(5,1+4)=5	[0,1,2,4,5,0,0,0]
10	w=5, i=0	5	(1,1)	dp[5] = max(dp[5], dp[5-1]+1) = max(0,5+1)=6	[0,1,2,4,5,6,0,0]
11	w=5, i=1	5	(3,4)	dp[5] = max(6, dp[5-3]+4) = max(6,2+4)=6	[0,1,2,4,5,6,0,0]
12	w=6, i=0	6	(1,1)	dp[6] = max(dp[6], dp[6-1]+1) = max(0,6+1)=7	[0,1,2,4,5,6,7,0]
13	w=6, i=1	6	(3,4)	dp[6] = max(7, dp[6-3]+4) = max(7,4+4)=8	[0,1,2,4,5,6,8,0]
14	w=7, i=0	7	(1,1)	dp[7] = max(dp[7], dp[7-1]+1) = max(0,8+1)=9	[0,1,2,4,5,6,8,9]
15	w=7, i=1	7	(3,4)	dp[7] = max(9, dp[7-3]+4) = max(9,5+4)=9	[0,1,2,4,5,6,8,9]
16	End	-	-	Final dp array computed	[0,1,2,4,5,6,8,9]

💡 Reached weight W=7, all items considered, dp[7] = 9 is max value

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 7	After Step 10	After Step 13	After Step 16
dp	[0,0,0,0,0,0,0,0]	[0,1,0,0,0,0,0,0]	[0,1,2,0,0,0,0,0]	[0,1,2,4,0,0,0,0]	[0,1,2,4,5,6,0,0]	[0,1,2,4,5,6,8,0]	[0,1,2,4,5,6,8,9]
w	0	1	2	3	5	6	7
i	-	0	0	1	0	1	-

Key Moments - 3 Insights

Why do we update dp[w] multiple times for the same weight w?

Because we consider all items for each weight w to find the best value. See execution_table rows 6 and 7 where dp[3] updates first with item 0 then with item 1.

Why can we use dp[w - item.weight] when updating dp[w]?

Because dp[w - item.weight] stores the best value for a smaller weight, allowing us to add the current item's value to build up dp[w]. This is shown in execution_table rows like 7 and 13.

Why does the dp array start with all zeros?

Because with weight 0, no items can be taken, so max value is 0. This initial state is shown in execution_table step 1.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the dp[3] value after considering item (3,4)?

A 3

B 0

C 4

D 7

💡 Hint

Check the DP Update column at step 7 in execution_table

At which step does dp[5] first get updated to 6?

A Step 10

B Step 11

C Step 9

D Step 12

💡 Hint

Look for dp[5] updates in execution_table rows 10 and 11

If the item (3,4) was removed, what would dp[7] be after all steps?

A 9

B 7

C 0

D 3

💡 Hint

Without item (3,4), only item (1,1) contributes, so dp[7] = 7 * 1 = 7, check variable_tracker dp final state

Concept Snapshot

Unbounded Knapsack Problem:
- Find max value with unlimited items
- Use dp array of size W+1
- For each weight w, check all items
- Update dp[w] = max(dp[w], dp[w - item.weight] + item.value)
- dp[W] holds max value
- Allows multiple uses of same item

Full Transcript

The Unbounded Knapsack Problem finds the maximum value achievable with unlimited copies of given items within a weight limit W. We use a dp array where dp[w] stores the best value for weight w. Starting from weight 1 to W, for each weight, we check all items. If an item's weight is less or equal to current weight w, we update dp[w] by comparing current dp[w] and dp[w - item.weight] plus the item's value. This way, we build up solutions from smaller weights to larger weights, allowing repeated use of items. The final answer is dp[W]. The execution table shows step-by-step how dp array updates for each weight and item. Key moments clarify why dp updates multiple times per weight and why dp starts with zeros. The visual quiz tests understanding of dp updates and final values.

Practice

1. What is the main difference between the Unbounded Knapsack Problem and the 0/1 Knapsack Problem?

easy

A. In Unbounded Knapsack, you can use unlimited copies of each item.

B. In Unbounded Knapsack, items cannot be repeated.

C. Unbounded Knapsack only works with fractional items.

D. Unbounded Knapsack has a fixed number of items to choose from.

Solution

1. Step 1: Understand item usage rules

   In 0/1 Knapsack, each item can be chosen at most once. In Unbounded Knapsack, items can be chosen unlimited times.

2. Step 2: Compare problem constraints

   Unbounded Knapsack allows multiple copies of the same item, unlike 0/1 Knapsack.

3. Final Answer:

   In Unbounded Knapsack, you can use unlimited copies of each item. -> Option A

4. Quick Check:

   Unlimited item usage = In Unbounded Knapsack, you can use unlimited copies of each item. [OK]

Hint: Remember: Unbounded means unlimited item copies allowed [OK]

Common Mistakes:

• Confusing Unbounded with fractional knapsack
• Thinking items can only be chosen once
• Assuming fixed number of items to pick

2. Which of the following C# code snippets correctly initializes the DP array for the Unbounded Knapsack problem of capacity W?

easy

A. int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0;

B. int[] dp = new int[W]; for (int i = 0; i < W; i++) dp[i] = -1;

C. int[] dp = new int[W + 1]; for (int i = 1; i <= W; i++) dp[i] = int.MaxValue;

D. int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = int.MinValue;

Solution

1. Step 1: Check array size and initialization

   The DP array size should be W+1 to cover capacities from 0 to W. Initializing with 0 is correct for max value problems.

2. Step 2: Validate initialization values

   Setting dp[i] = 0 means no value at start, which is correct. Negative or max int values are incorrect here.

3. Final Answer:

   int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0; -> Option A

4. Quick Check:

   DP array size W+1 and zero initialization = int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0; [OK]

Hint: DP array size = capacity + 1, initialize with zero for max value [OK]

Common Mistakes:

• Using wrong array size (W instead of W+1)
• Initializing with int.MaxValue or int.MinValue incorrectly
• Starting loop from 1 instead of 0

3. What is the output of the following C# code snippet for Unbounded Knapsack?

int W = 5;
int[] wt = {1, 3, 4};
int[] val = {10, 40, 50};
int[] dp = new int[W + 1];
for (int i = 0; i <= W; i++) {
    for (int j = 0; j < wt.Length; j++) {
        if (wt[j] <= i) {
            dp[i] = Math.Max(dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}
Console.WriteLine(dp[W]);

medium

A. 120

B. 100

C. 60

D. 90

Solution

1. Step 1: Trace dp array updates for each capacity

   Calculate dp values: - dp[0]=0 - dp[1]=max(dp[1], dp[0]+10)=10 - dp[2]=max(dp[2], dp[1]+10)=20 - dp[3]=max(dp[3], dp[0]+40, dp[2]+10)=max(0,40,30)=40 - dp[4]=max(dp[4], dp[1]+40, dp[3]+10, dp[0]+50)=max(0,50,50,50)=50 - dp[5]=max(dp[5], dp[2]+40, dp[4]+10, dp[1]+50)=max(0,60,60,60)=60 But re-check dp[5] carefully: dp[5] = max(dp[5], dp[5-1]+10=dp[4]+10=50+10=60, dp[5-3]+40=dp[2]+40=20+40=60, dp[5-4]+50=dp[1]+50=10+50=60) = 60

2. Step 2: Recalculate carefully for maximum value

   Check if multiple items can be combined: - For dp[5], try dp[5-1]+10=dp[4]+10=50+10=60 - For dp[4], dp[4-4]+50=dp[0]+50=50 - For dp[3], dp[3-3]+40=dp[0]+40=40 - For dp[2], dp[2-1]+10=dp[1]+10=10+10=20 - For dp[1], dp[1-1]+10=dp[0]+10=10 So dp[5] = 60

3. Final Answer:

   60 -> Option C

4. Quick Check:

   Maximum value for capacity 5 = 60 [OK]

Hint: Trace dp array from 0 to W, update max values [OK]

Common Mistakes:

• Not considering multiple copies of items
• Incorrect dp array indexing
• Mixing up weights and values

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

int W = 7;
int[] wt = {1, 3, 4};
int[] val = {10, 40, 50};
int[] dp = new int[W + 1];
for (int i = 0; i < W; i++) {
    for (int j = 0; j < wt.Length; j++) {
        if (wt[j] <= i) {
            dp[i] = Math.Max(dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}
Console.WriteLine(dp[W]);

medium

A. The inner loop should run from 1 to wt.Length.

B. The condition should be wt[j] < i instead of wt[j] <= i.

C. The dp array size should be W instead of W + 1.

D. The outer loop should run from 0 to W inclusive (i <= W).

Solution

1. Step 1: Check loop ranges for dp array access

   The dp array has size W+1, so valid indices are 0 to W inclusive. The outer loop runs only to i < W, missing dp[W].

2. Step 2: Identify impact of loop range

   Because dp[W] is not computed, the final answer dp[W] will be incorrect or default zero.

3. Final Answer:

   The outer loop should run from 0 to W inclusive (i <= W). -> Option D

4. Quick Check:

   Loop must cover full dp array size = The outer loop should run from 0 to W inclusive (i <= W). [OK]

Hint: Loop dp array fully from 0 to W inclusive [OK]

Common Mistakes:

• Using wrong loop bounds causing incomplete dp calculation
• Incorrect dp array size
• Wrong comparison operators in conditions

5. Given items with weights [2, 3, 5] and values [50, 70, 100], and a knapsack capacity of 8, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 270

B. 200

C. 250

D. 220

Solution

1. Step 1: Calculate dp array values for capacities 0 to 8

   Initialize dp[0]=0. - dp[1]=0 (no item fits) - dp[2]=max(dp[2], dp[0]+50)=50 - dp[3]=max(dp[3], dp[0]+70)=70 - dp[4]=max(dp[4], dp[2]+50=50+50=100, dp[1]+70=0+70=70)=100 - dp[5]=max(dp[5], dp[0]+100=100, dp[3]+50=70+50=120, dp[2]+70=50+70=120)=120 - dp[6]=max(dp[6], dp[4]+50=100+50=150, dp[3]+70=70+70=140, dp[1]+100=0+100=100)=150 - dp[7]=max(dp[7], dp[5]+50=120+50=170, dp[4]+70=100+70=170, dp[2]+100=50+100=150)=170 - dp[8]=max(dp[8], dp[6]+50=150+50=200, dp[5]+70=120+70=190, dp[3]+100=70+100=170)=200

2. Step 2: Check for better combinations

   Try combining items differently: - Using four 2-weight items: 4*50=200 - Using two 3-weight + one 2-weight: 2*70+50=190 - Using one 5-weight + one 3-weight: 100+70=170 - Using one 5-weight + one 3-weight + one 2-weight (over capacity) No better than 200 found, but check dp[8] carefully: - dp[8] = max(200, 190, 170) = 200

3. Step 3: Re-examine for missed combinations

   Try dp[8] = dp[3] + dp[5] = 70 + 120 = 190 (less than 200) Try dp[8] = dp[2] + dp[6] = 50 + 150 = 200 Try dp[8] = dp[0] + 100 (weight 5) + dp[3] (weight 3) = 100 + 70 = 170 Try dp[8] = dp[4] + dp[4] = 100 + 100 = 200 Try dp[8] = dp[7] + dp[1] = 170 + 0 = 170 Try dp[8] = dp[6] + dp[2] = 150 + 50 = 200 Try dp[8] = dp[5] + dp[3] = 120 + 70 = 190 Try dp[8] = dp[8-5] + 100 = dp[3] + 100 = 70 + 100 = 170 Try dp[8] = dp[8-3] + 70 = dp[5] + 70 = 120 + 70 = 190 Try dp[8] = dp[8-2] + 50 = dp[6] + 50 = 150 + 50 = 200 No combination exceeds 200, so 200 is max.

4. Final Answer:

   200 -> Option B

5. Quick Check:

   Max value for capacity 8 = 200 [OK]

Hint: Build dp array stepwise, combine items for max value [OK]

Common Mistakes:

• Ignoring multiple copies of items
• Not checking all item combinations
• Incorrect dp array updates