0 1 Knapsack Problem in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the 0 1 Knapsack Problem?

To select items with given weights and values to maximize total value without exceeding the knapsack's weight capacity.

beginner

In the 0 1 Knapsack Problem, what does the '0 1' mean?

Each item can be either taken (1) or not taken (0); partial items are not allowed.

intermediate

What is the time complexity of the dynamic programming solution for the 0 1 Knapsack Problem?

O(n * W), where n is the number of items and W is the maximum weight capacity of the knapsack.

intermediate

What does the DP table represent in the 0 1 Knapsack Problem?

It stores the maximum value achievable using the first i items with a weight limit j.

intermediate

How do you decide whether to include an item in the knapsack during the DP solution?

Compare the value of including the item (value + best value for remaining weight) with excluding it, and choose the maximum.

What does the 'weight capacity' in the 0 1 Knapsack Problem represent?

A Total number of items available

B Maximum total weight the knapsack can hold

C Maximum value achievable

D Number of items selected

The weight capacity is the limit on the total weight the knapsack can carry.

Which approach is commonly used to solve the 0 1 Knapsack Problem efficiently?

A Dynamic Programming

B Greedy Algorithm

C Brute Force without optimization

D Divide and Conquer

Dynamic Programming efficiently solves the problem by storing intermediate results.

In the DP table, what does the entry dp[i][j] represent?

A Maximum value using first i items with weight limit j

B Weight of the ith item

C Number of items selected so far

D Minimum weight to achieve value j

dp[i][j] stores the best value achievable with first i items and weight limit j.

If an item's weight is more than the current weight limit j, what happens in the DP solution?

A Weight limit j is increased

B Item is included anyway

C Algorithm stops

D Item is excluded; dp[i][j] = dp[i-1][j]

If the item is too heavy, it cannot be included, so the value remains as without it.

What is the key difference between 0 1 Knapsack and Fractional Knapsack?

A 0 1 Knapsack allows partial items; Fractional does not

B Both allow partial items

C 0 1 Knapsack does not allow partial items; Fractional does

D Neither allows partial items

0 1 Knapsack requires whole items; Fractional Knapsack allows fractions of items.

Explain how dynamic programming solves the 0 1 Knapsack Problem step-by-step.

Think about building solutions from smaller subproblems.

Describe the difference between the 0 1 Knapsack Problem and the Fractional Knapsack Problem.

Focus on item selection rules and solution methods.

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