Unbounded Knapsack Problem in DSA Rust - 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

↓

Result is DP[W

↓

End

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

Execution Sample

DSA Rust

let weights = [1, 3, 4];
let values = [1, 4, 5];
let W = 7;
let mut dp = vec![0; W + 1];
for w in 1..=W {
    for i in 0..weights.len() {
        if weights[i] <= w {
            dp[w] = dp[w].max(dp[w - weights[i]] + values[i]);
        }
    }
}
println!("{}", dp[W]);

Calculates max value for knapsack capacity 7 with items (1,1), (3,4), (4,5) allowing unlimited item use.

Execution Table

Step	Current Weight (w)	Item Index (i)	Condition (weights[i] <= w)	DP Before	DP Update	DP After	DP Array State
1	1	0	1 <= 1: True	dp[1]=0	max(0, dp[1-1]+1)=max(0,0+1)=1	1	[0,1,0,0,0,0,0,0]
2	1	1	3 <= 1: False	dp[1]=1	No update	1	[0,1,0,0,0,0,0,0]
3	1	2	4 <= 1: False	dp[1]=1	No update	1	[0,1,0,0,0,0,0,0]
4	2	0	1 <= 2: True	dp[2]=0	max(0, dp[2-1]+1)=max(0,1+1)=2	2	[0,1,2,0,0,0,0,0]
5	2	1	3 <= 2: False	dp[2]=2	No update	2	[0,1,2,0,0,0,0,0]
6	2	2	4 <= 2: False	dp[2]=2	No update	2	[0,1,2,0,0,0,0,0]
7	3	0	1 <= 3: True	dp[3]=0	max(0, dp[3-1]+1)=max(0,2+1)=3	3	[0,1,2,3,0,0,0,0]
8	3	1	3 <= 3: True	dp[3]=3	max(3, dp[3-3]+4)=max(3,0+4)=4	4	[0,1,2,4,0,0,0,0]
9	3	2	4 <= 3: False	dp[3]=4	No update	4	[0,1,2,4,0,0,0,0]
10	4	0	1 <= 4: True	dp[4]=0	max(0, dp[4-1]+1)=max(0,4+1)=5	5	[0,1,2,4,5,0,0,0]
11	4	1	3 <= 4: True	dp[4]=5	max(5, dp[4-3]+4)=max(5,1+4)=5	5	[0,1,2,4,5,0,0,0]
12	4	2	4 <= 4: True	dp[4]=5	max(5, dp[4-4]+5)=max(5,0+5)=5	5	[0,1,2,4,5,0,0,0]
13	5	0	1 <= 5: True	dp[5]=0	max(0, dp[5-1]+1)=max(0,5+1)=6	6	[0,1,2,4,5,6,0,0]
14	5	1	3 <= 5: True	dp[5]=6	max(6, dp[5-3]+4)=max(6,2+4)=6	6	[0,1,2,4,5,6,0,0]
15	5	2	4 <= 5: True	dp[5]=6	max(6, dp[5-4]+5)=max(6,1+5)=6	6	[0,1,2,4,5,6,0,0]
16	6	0	1 <= 6: True	dp[6]=0	max(0, dp[6-1]+1)=max(0,6+1)=7	7	[0,1,2,4,5,6,7,0]
17	6	1	3 <= 6: True	dp[6]=7	max(7, dp[6-3]+4)=max(7,4+4)=8	8	[0,1,2,4,5,6,8,0]
18	6	2	4 <= 6: True	dp[6]=8	max(8, dp[6-4]+5)=max(8,2+5)=8	8	[0,1,2,4,5,6,8,0]
19	7	0	1 <= 7: True	dp[7]=0	max(0, dp[7-1]+1)=max(0,8+1)=9	9	[0,1,2,4,5,6,8,9]
20	7	1	3 <= 7: True	dp[7]=9	max(9, dp[7-3]+4)=max(9,5+4)=9	9	[0,1,2,4,5,6,8,9]
21	7	2	4 <= 7: True	dp[7]=9	max(9, dp[7-4]+5)=max(9,4+5)=9	9	[0,1,2,4,5,6,8,9]

💡 Completed all weights from 1 to 7; dp[7] = 9 is the max value.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
dp[0]	0	0	0	0	0	0	0	0	0
dp[1]	0	1	1	1	1	1	1	1	1
dp[2]	0	0	2	2	2	2	2	2	2
dp[3]	0	0	0	4	4	4	4	4	4
dp[4]	0	0	0	0	5	5	5	5	5
dp[5]	0	0	0	0	0	6	6	6	6
dp[6]	0	0	0	0	0	0	8	8	8
dp[7]	0	0	0	0	0	0	0	9	9

Key Moments - 3 Insights

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

Because we check all items for each weight w, dp[w] is updated to the max value possible by including any item multiple times, as shown in execution_table rows where dp[w] changes after considering different items.

Why do we use dp[w - weights[i]] + values[i] when updating dp[w]?

This represents taking the current item i once plus the best value for the remaining weight w - weights[i]. This allows unlimited use because dp[w - weights[i]] already includes optimal solutions possibly using item i multiple times, as seen in the execution_table updates.

Why does the dp array start with zeros and get updated from smaller weights to larger?

We build solutions from smaller weights to larger so that when we compute dp[w], dp[w - weights[i]] is already computed and correct. This bottom-up approach is visible in the execution_table where dp values for smaller weights are used to update larger weights.

Visual Quiz - 3 Questions

Look at the execution_table at step 17, what is the dp[6] value after update?

A 7

B 8

C 6

D 9

💡 Hint

Check the DP After column for step 17 in the execution_table.

At which current weight w does dp[w] first reach the value 5?

A w = 3

B w = 5

C w = 4

D w = 6

💡 Hint

Look at the DP Array State column in execution_table rows for weights 3 and 4.

If the item with weight 3 and value 4 was removed, how would dp[7] change?

A dp[7] would be 8

B dp[7] would be 9

C dp[7] would be 5

D dp[7] would be 6

💡 Hint

Without the item (3,4), dp updates rely on items (1,1) and (4,5). Check variable_tracker dp[7] values and reasoning.

Concept Snapshot

Unbounded Knapsack Problem:
- Use a DP array dp of size W+1 initialized to 0.
- For each weight w from 1 to W:
  - For each item i:
    - If weights[i] <= w, update dp[w] = max(dp[w], dp[w - weights[i]] + values[i])
- dp[W] holds the max value allowing unlimited item use.
- Bottom-up approach ensures smaller subproblems solved first.

Full Transcript

The Unbounded Knapsack Problem is solved using a dynamic programming array dp where dp[w] represents the maximum value achievable with capacity w. We start with dp initialized to zero. For each weight from 1 to W, we check each item. If the item's weight is less than or equal to the current weight, we consider including it by updating dp[w] to the maximum of its current value and the value of dp[w - item weight] plus the item's value. This allows unlimited use of items because dp[w - item weight] already includes optimal solutions possibly using the same item multiple times. The process continues until dp[W] is computed, which gives the maximum value for the full capacity. The execution table shows step-by-step updates of dp array values, and the variable tracker records dp values after each weight iteration. Key moments clarify why dp[w] updates multiple times, why dp[w - weights[i]] + values[i] is used, and why the dp array is built from smaller to larger weights. The visual quiz tests understanding of dp values at specific steps and the effect of removing an item. The concept snapshot summarizes the approach in a few lines.

Practice

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

easy

A. There is no difference; both problems are the same.

B. In Unbounded Knapsack, items cannot be repeated.

C. In 0/1 Knapsack, items can be used unlimited times.

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

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 multiple times.

2. Step 2: Compare problem definitions

   Unbounded Knapsack allows unlimited copies of each item, unlike 0/1 Knapsack.

3. Final Answer:

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

4. Quick Check:

   Unbounded = unlimited copies [OK]

Hint: Unbounded means unlimited items allowed [OK]

Common Mistakes:

• Confusing 0/1 and unbounded item usage
• Thinking items cannot repeat in unbounded knapsack
• Assuming both problems have same constraints

2. Which of the following Rust code snippets correctly initializes the DP array for the Unbounded Knapsack problem with capacity `W`?

easy

A. let mut dp = vec![std::i32::MAX; W + 1];

B. let mut dp = vec![1; W];

C. let mut dp = vec![0; W + 1];

D. let mut dp = vec![None; W + 1];

Solution

1. Step 1: Understand DP array purpose

   DP array stores max value for capacities from 0 to W, initialized with 0 as base case.

2. Step 2: Check initialization correctness

   let mut dp = vec![0; W + 1]; initializes dp with zeros for all capacities including 0 to W, which is correct.

3. Final Answer:

   let mut dp = vec![0; W + 1]; -> Option C

4. Quick Check:

   DP array zero-initialized for capacities [OK]

Hint: DP array size = capacity + 1, initialize with zeros [OK]

Common Mistakes:

• Using MAX instead of 0 for initialization
• Wrong vector size (W instead of W+1)
• Using None instead of integer values

3. What will be the output of the following Rust code snippet for Unbounded Knapsack?

let weights = vec![2, 3, 4];
let values = vec![15, 20, 30];
let capacity = 7;
let mut dp = vec![0; capacity + 1];

for w in 1..=capacity {
    for i in 0..weights.len() {
        if weights[i] <= w {
            dp[w] = dp[w].max(dp[w - weights[i]] + values[i]);
        }
    }
}
println!("{}", dp[capacity]);

medium

A. 65

B. 50

C. 45

D. 60

Solution

1. Step 1: Trace DP array updates for capacity 7

   Calculate max value for each capacity: - w=1: no items fit, dp[1]=0 - w=2: use item 0 (weight 2, value 15), dp[2]=15 - w=3: item 1 fits (3,20), dp[3]=20 - w=4: item 2 fits (4,30), dp[4]=30; also dp[4-2]=dp[2]=15 + 15=30, max=30 - w=5: check items: item0: dp[5-2]=dp[3]=20 +15=35 item1: dp[5-3]=dp[2]=15 +20=35 item2: dp[5-4]=dp[1]=0 +30=30 max=35 - w=6: item0: dp[6-2]=dp[4]=30 +15=45 item1: dp[6-3]=dp[3]=20 +20=40 item2: dp[6-4]=dp[2]=15 +30=45 max=45 - w=7: item0: dp[7-2]=dp[5]=35 +15=50 item1: dp[7-3]=dp[4]=30 +20=50 item2: dp[7-4]=dp[3]=20 +30=50 max=50

2. Step 2: Verify final dp[7]

   dp[7] = 50

3. Final Answer:

   50 -> Option B

4. Quick Check:

   Max value at capacity 7 = 50 [OK]

Hint: Check all items for each capacity and pick max value [OK]

Common Mistakes:

• Not considering multiple uses of same item
• Incorrectly updating dp array
• Off-by-one errors in loops

4. Identify the bug in the following Rust code snippet for Unbounded Knapsack and select the correct fix:

let weights = vec![1, 3, 4];
let values = vec![10, 40, 50];
let capacity = 6;
let mut dp = vec![0; capacity];

for w in 1..=capacity {
    for i in 0..weights.len() {
        if weights[i] <= w {
            dp[w] = dp[w].max(dp[w - weights[i]] + values[i]);
        }
    }
}
println!("{}", dp[capacity]);

medium

A. Initialize dp with size capacity + 1 instead of capacity.

B. Change loop to for w in 0..capacity instead of 1..=capacity.

C. Use dp[w + weights[i]] instead of dp[w - weights[i]].

D. Replace max with min in dp update.

Solution

1. Step 1: Check dp vector size

   dp is initialized with size capacity (6), but indices go up to capacity (6), so dp[6] access is out of bounds.

2. Step 2: Fix dp size to capacity + 1

   Initialize dp with size capacity + 1 to safely access dp[capacity].

3. Final Answer:

   Initialize dp with size capacity + 1 instead of capacity. -> Option A

4. Quick Check:

   dp size = capacity + 1 to avoid index error [OK]

Hint: DP array size must be capacity + 1 for indexing [OK]

Common Mistakes:

• Off-by-one error in dp array size
• Accessing dp out of bounds
• Incorrect loop ranges

5. Given items with weights [1, 3, 4] and values [10, 40, 50], and a knapsack capacity of 7, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 90

B. 120

C. 140

D. 110

Solution

1. Step 1: Calculate dp values for capacities 1 to 7

   Weights = [1,3,4], Values = [10,40,50] - w=1: dp[1] = max(dp[1-1]+10) = 10 - w=2: dp[2] = max(dp[2-1]+10) = dp[1]+10=20 - w=3: dp[3] = max(dp[3-1]+10=dp[2]+10=30, dp[3-3]+40=dp[0]+40=40) = 40 - w=4: dp[4] = max(dp[4-1]+10=dp[3]+10=50, dp[4-3]+40=dp[1]+40=50, dp[4-4]+50=dp[0]+50=50) = 50 - w=5: dp[5] = max(dp[5-1]+10=dp[4]+10=60, dp[5-3]+40=dp[2]+40=60, dp[5-4]+50=dp[1]+50=60) = 60 - w=6: dp[6] = max(dp[6-1]+10=dp[5]+10=70, dp[6-3]+40=dp[3]+40=80, dp[6-4]+50=dp[2]+50=70) = 80 - w=7: dp[7] = max(dp[7-1]+10=dp[6]+10=90, dp[7-3]+40=dp[4]+40=90, dp[7-4]+50=dp[3]+50=90) = 90

2. Step 2: Re-check calculations for maximum value

   Check if combining items differently yields more: Try using item 3 (weight 4, value 50) twice: total weight 8 > 7 (not possible) Try using item 1 (weight 1, value 10) seven times: 7*10=70 < 90 Try using item 2 (weight 3, value 40) twice + item 1 once: 3+3+1=7 weight, value=40+40+10=90 Maximum value is 90.

3. Final Answer:

   90 -> Option A

4. Quick Check:

   Max value at capacity 7 = 90 [OK]

Hint: Build dp from 1 to capacity, pick max value each step [OK]

Common Mistakes:

• Ignoring multiple item usage
• Wrong dp updates
• Miscalculating max values