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

Time Complexity: Unbounded Knapsack Problem

O(W * n)

Understanding Time Complexity

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

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 n, int val[], int wt[]) {
    int dp[W+1] = {0};
    for (int i = 0; i <= W; i++) {
        for (int j = 0; j < n; j++) {
            if (wt[j] <= i) {
                dp[i] = std::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 copies of n items.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop runs from 0 to W and the inner loop runs over all n items.
• How many times: Outer loop runs W+1 times, inner loop runs n times for each outer iteration, so total roughly (W+1)*n times.

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 (W * n)
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(W * n)

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

Common Mistake

[X] Wrong: "The time complexity is exponential because we can pick unlimited items."

[OK] Correct: The dynamic programming approach avoids repeated work by storing results for each capacity, so it runs in polynomial time, not exponential.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency over naive methods, a key skill in problem solving.

Self-Check

"What if we changed the problem to allow only one copy of each item? 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, items cannot be repeated.

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

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

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

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 definitions

   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 B

4. Quick Check:

   Unbounded Knapsack = unlimited item usage [OK]

Hint: Remember: Unbounded means unlimited items allowed [OK]

Common Mistakes:

• Confusing 0/1 Knapsack with Unbounded Knapsack
• Thinking items cannot repeat in Unbounded Knapsack
• Assuming both problems have same constraints

2. Which of the following is the correct way to initialize the DP array for the Unbounded Knapsack problem in C++?

easy

A. int dp[W+1]; fill(dp, dp+W+1, 0);

B. int dp[W]; // uninitialized array

C. int dp[W+1]; fill(dp, dp+W+1, -1);

D. int dp[W+1] = {0}; // W is max weight

Solution

1. Step 1: Understand DP array role

   DP array stores max value for weights from 0 to W. It should start with 0 for no items chosen.

2. Step 2: Choose correct initialization

   Filling dp with 0 means no value at start, which is correct. int dp[W+1]; fill(dp, dp+W+1, 0); uses fill with 0 correctly.

3. Final Answer:

   int dp[W+1]; fill(dp, dp+W+1, 0); -> Option A

4. Quick Check:

   DP array initialized to 0 for all weights [OK]

Hint: Initialize DP with zeros for max value calculations [OK]

Common Mistakes:

• Leaving dp uninitialized causing garbage values
• Initializing dp with -1 which is incorrect here
• Using wrong array size causing out-of-bound errors

3. What will be the output of the following C++ code snippet for Unbounded Knapsack with weights = {2, 3}, values = {10, 15}, and max weight W = 5?

int weights[] = {2, 3};
int values[] = {10, 15};
int W = 5;
int dp[6] = {0};
for (int i = 0; i <= W; i++) {
  for (int j = 0; j < 2; j++) {
    if (weights[j] <= i) {
      dp[i] = max(dp[i], dp[i - weights[j]] + values[j]);
    }
  }
}
std::cout << dp[W] << std::endl;

medium

A. 30

B. 35

C. 25

D. 40

Solution

1. Step 1: Trace dp array updates for weights 0 to 5

   Calculate dp[i]: dp[0]=0 For i=1: dp[1]=0 (no items fit) For i=2: dp[2]=max(dp[2], dp[0]+10)=10 For i=3: dp[3]=max(dp[3], dp[0]+15)=15 For i=4: dp[4]=max(dp[4], dp[2]+10=20, dp[1]+15=15)=20 For i=5: dp[5]=max(dp[5], dp[3]+10=25, dp[2]+15=25) so dp[5]=25

2. Step 2: Confirm final dp[5] value

   Maximum value for weight 5 is 25 by choosing items (3 + 2) or (2 + 3).

3. Final Answer:

   25 -> Option C

4. Quick Check:

   Max value at W=5 is 25 [OK]

Hint: Check dp updates for each weight carefully [OK]

Common Mistakes:

• Adding values incorrectly for dp updates
• Ignoring multiple item usage possibilities
• Mixing up indices in dp array

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

int dp[W+1];
for (int i = 0; i <= W; i++) {
  for (int j = 0; j < n; j++) {
    if (weights[j] <= i) {
      dp[i] = max(dp[i], dp[i - weights[j]] + values[j]);
    }
  }
}

medium

A. dp array is not initialized before use.

B. Loop indices are incorrect.

C. Weights and values arrays are not declared.

D. The max function is used incorrectly.

Solution

1. Step 1: Check dp array initialization

   dp array is declared but not initialized, so it contains garbage values affecting max calculations.

2. Step 2: Importance of initializing dp

   dp should be initialized to 0 for all indices before the loops to ensure correct max value updates.

3. Final Answer:

   dp array is not initialized before use. -> Option A

4. Quick Check:

   Uninitialized dp causes wrong results [OK]

Hint: Always initialize dp array before loops [OK]

Common Mistakes:

• Forgetting to initialize dp array
• Assuming dp defaults to zero
• Ignoring garbage values in dp

5. Given items with weights = {1, 3, 4}, values = {15, 50, 60}, and max weight W = 7, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 105

B. 120

C. 150

D. 115

Solution

1. Step 1: Calculate dp values from 0 to 7

   dp[0]=0 For i=1: dp[1]=max(0, dp[0]+15)=15 For i=2: dp[2]=max(0, dp[1]+15)=30 For i=3: dp[3]=max(0, dp[0]+50, dp[2]+15)=50 For i=4: dp[4]=max(0, dp[1]+50, dp[0]+60, dp[3]+15)=65 For i=5: dp[5]=max(0, dp[2]+50, dp[1]+60, dp[4]+15)=80 For i=6: dp[6]=max(0, dp[3]+50, dp[2]+60, dp[5]+15)=100 For i=7: dp[7]=max(0, dp[4]+50=115, dp[3]+60=110, dp[6]+15=115)=115

2. Step 2: Confirm maximum value at W=7

   The maximum value achievable is 115 by combining items optimally.

3. Final Answer:

   115 -> Option D

4. Quick Check:

   Max value at W=7 is 115 [OK]

Hint: Build dp array stepwise for each weight [OK]

Common Mistakes:

• Not considering all item combinations
• Incorrect dp updates for each weight
• Mixing up weights and values indices