0 1 Knapsack Problem in DSA C - Time & Space Complexity

Time Complexity: 0 1 Knapsack Problem

O(n * W)

Understanding Time Complexity

We want to understand how the time needed to solve the 0 1 Knapsack Problem changes as the number of items and the knapsack capacity grow.

How does the algorithm's work increase when we add more items or increase the capacity?

Scenario Under Consideration

Analyze the time complexity of the following dynamic programming solution.

int max(int a, int b) {
    return (a > b) ? a : b;
}

int knapsack(int W, int wt[], int val[], int n) {
    int dp[n+1][W+1];
    for (int i = 0; i <= n; i++) {
        for (int w = 0; w <= W; w++) {
            if (i == 0 || w == 0)
                dp[i][w] = 0;
            else if (wt[i-1] <= w)
                dp[i][w] = max(val[i-1] + dp[i-1][w - wt[i-1]], dp[i-1][w]);
            else
                dp[i][w] = dp[i-1][w];
        }
    }
    return dp[n][W];
}
    

This code fills a table to find the best value achievable with given items and capacity.

Identify Repeating Operations

Look at the loops that repeat work:

• Primary operation: The nested loops over items (n) and capacity (W).
• How many times: The outer loop runs n+1 times, and the inner loop runs W+1 times for each outer loop.

How Execution Grows With Input

As we increase the number of items or the capacity, the work grows by multiplying these two.

Input Size (n, W)	Approx. Operations
10 items, capacity 10	~100 operations
100 items, capacity 100	~10,000 operations
1000 items, capacity 1000	~1,000,000 operations

Pattern observation: Doubling items and capacity roughly squares the work needed.

Final Time Complexity

Time Complexity: O(n * W)

This means the time grows proportionally to the number of items times the capacity size.

Common Mistake

[X] Wrong: "The time depends only on the number of items, not the capacity."

[OK] Correct: The capacity size directly affects how many subproblems we solve, so it impacts the total work.

Interview Connect

Understanding this complexity helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts.

Self-Check

"What if we used a recursive solution without memoization? How would the time complexity change?"

Practice

1. In the 0/1 Knapsack problem, why can't we take fractions of items like in the Fractional Knapsack?

easy

A. Because the DP table only stores integers

B. Because each item must be taken as a whole unit or not at all - fractional inclusion violates the problem definition

C. Because fractions make the recursion infinite

D. Because the weight array only accepts integer values

Solution

1. 0/1 Constraint

   The '0/1' in the name literally means each item has two states: not taken (0) or fully taken (1). There is no state for 0.5 or any fraction.

2. Why Greedy Fails

   Because you cannot take fractions, the greedy 'highest value/weight ratio first' strategy fails - taking a high-ratio item may block a better combination of whole items.

3. Why DP is Needed

   DP explores all combinations of 0/1 choices, which is required when greedy fails due to the binary selection constraint.

4. Final Answer:

   Each item is binary - take it whole or skip it. → Option B

Hint: 0/1 = binary choice per item: take it whole or skip entirely. [OK]

Common Mistakes:

• Confusing 0/1 Knapsack with Fractional Knapsack
• Thinking DP table data type restricts selection
• Assuming greedy works for 0/1 Knapsack

2. For 0/1 Knapsack with n=3 items and capacity W=5, what is the size of the standard DP table dp[n+1][W+1] in C?

easy

A. 3 × 5

B. 4 × 5

C. 3 × 6

D. 4 × 6

Solution

1. Table Dimensions

   The DP table is dp[n+1][W+1]. With n=3 items and W=5 capacity: dp[4][6].

2. Why +1

   Index 0 represents the base case: dp[0][w] = 0 (no items) and dp[i][0] = 0 (zero capacity). The +1 makes these base cases fit naturally without off-by-one adjustments.

3. In C

   int dp[4][6]; - 4 rows (0 to 3 items), 6 columns (capacity 0 to 5).

4. Final Answer:

   4 × 6 → Option D

Hint: dp[n+1][W+1] → always add 1 to both dimensions for the base case. [OK]

Common Mistakes:

• Using dp[n][W] and missing the base case row/column
• Adding +1 to only one dimension
• Confusing n items with n+1 rows

3. In the 0/1 Knapsack DP recurrence, when weight[i] > w (current item is heavier than current capacity), what is dp[i][w]?

medium

A. 0

B. dp[i-1][w] - same value as without this item

C. dp[i][w-1]

D. value[i] + dp[i-1][w - weight[i]]

Solution

1. Can't Include Overweight Item

   If weight[i] > w, item i cannot fit - we must skip it.

2. Inherit from Previous Row

   dp[i][w] = dp[i-1][w]: the best value with first i items equals the best with first i-1 items.

3. Full Recurrence

   if (weight[i] > w): dp[i][w] = dp[i-1][w]. else: dp[i][w] = max(dp[i-1][w], value[i] + dp[i-1][w-weight[i]])

4. Final Answer:

   dp[i-1][w] → Option B

Hint: Item too heavy → skip → copy from row above: dp[i-1][w]. [OK]

Common Mistakes:

• Setting dp[i][w]=0 instead of carrying forward the previous best
• Applying include formula when item doesn't fit
• Reading dp[i][w-1] instead of dp[i-1][w]

4. What is the time and space complexity of the standard bottom-up 0/1 Knapsack DP solution?

medium

A. Time O(n×W), Space O(n×W)

B. Time O(n×W), Space O(W)

C. Time O(2^n), Space O(n)

D. Time O(n log W), Space O(n×W)

Solution

1. Time Complexity

   Two nested loops: outer over n items, inner over W+1 capacities. Each cell O(1). Total: O(n×W).

2. Space Complexity

   Full table dp[n+1][W+1] → O(n×W) space.

3. Note

   Space can be reduced to O(W) with 1D rolling array, but standard 2D table is O(n×W).

4. Final Answer:

   Time O(n×W), Space O(n×W) → Option A

Hint: Standard 2D table = O(n×W) time AND space. 1D optimized = O(W) space only. [OK]

Common Mistakes:

• Confusing standard O(n×W) space with optimized O(W)
• Claiming O(2^n) which applies only to naive recursion
• Mixing up time and space complexities

5. In the space-optimized 1D 0/1 Knapsack in C, why must the inner capacity loop iterate from W down to weight[i], not from 0 up to W?

hard

A. To avoid negative array indices

B. To prevent using item i more than once - forward iteration would allow the same item to be counted multiple times in the same pass

C. Because C arrays are stored in column-major order

D. To match the order of the weight array

Solution

1. Problem With Left-to-Right

   With a 1D array, processing left to right means dp[w-weight[i]] has already been updated in this pass - item i gets counted again at smaller capacities.

2. Right-to-Left Prevents Reuse

   Iterating from W down to weight[i] ensures dp[w-weight[i]] still holds the pre-item-i value. Each item counted at most once.

3. 0/1 vs Unbounded

   Right-to-left = 0/1 Knapsack. Left-to-right = Unbounded Knapsack.

4. Final Answer:

   Prevents reusing item i in the same pass. → Option B

Hint: Right-to-left = 0/1 (no reuse). Left-to-right = Unbounded (reuse allowed). [OK]

Common Mistakes:

• Using left-to-right which implements Unbounded Knapsack accidentally
• Thinking loop direction is about array bounds not item reuse
• Not knowing right-to-left is the key 0/1 vs Unbounded distinction