Unbounded Knapsack Problem in DSA C# - Time & Space Complexity

Time Complexity: Unbounded Knapsack Problem

O(n * W)

Understanding Time Complexity

We want to understand how the time needed to solve the Unbounded Knapsack problem grows as the input size increases.

Specifically, how the number of items and the knapsack capacity affect the work done.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int UnboundedKnapsack(int W, int[] wt, int[] val, int n) {
    int[] dp = new int[W + 1];
    for (int i = 0; i <= W; i++) {
        for (int j = 0; j < n; j++) {
            if (wt[j] <= i) {
                dp[i] = Math.Max(dp[i], dp[i - wt[j]] + val[j]);
            }
        }
    }
    return dp[W];
}
    

This code finds the maximum value for a knapsack of capacity W using unlimited items from n types.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops: outer loop runs from 0 to W, inner loop runs over n items.
• How many times: Outer loop runs W+1 times, inner loop runs n times each outer iteration.

How Execution Grows With Input

As the knapsack capacity W grows, the outer loop runs more times. For each capacity, we check all n items.

Input Size (W)	Approx. Operations (n * W)
10	10 * n
100	100 * n
1000	1000 * n

Pattern observation: The work grows linearly with both the knapsack capacity and the number of items.

Final Time Complexity

Time Complexity: O(n * W)

This means the time needed grows proportionally with the number of items and the knapsack capacity.

Common Mistake

[X] Wrong: "Since items can be chosen unlimited times, the time complexity is exponential because of infinite choices."

[OK] Correct: The dynamic programming approach uses a table to store results for each capacity, avoiding repeated work, so the complexity is polynomial, not exponential.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming efficiently solves problems with repeated choices, a key skill in interviews.

Self-Check

"What if we used recursion without memoization? How would the time complexity change?"

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