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

Recall & Review

beginner

What does the '0 1' in the 0 1 Knapsack Problem mean?

It means each item can either be taken completely (1) or not taken at all (0). You cannot take fractional parts of an item.

beginner

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

To maximize the total value of items put into the knapsack without exceeding its weight capacity.

intermediate

In the 0 1 Knapsack Problem, what does the dynamic programming table represent?

It stores the maximum value achievable with the first i items and a knapsack capacity of j.

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 knapsack capacity.

intermediate

Why can't greedy algorithms solve the 0 1 Knapsack Problem optimally?

Because taking the item with the best value or value-to-weight ratio first may prevent selecting a combination of items that yields a higher total value.

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

A Taking the whole item

B Taking half the item

C Not taking the item

D Taking multiple copies of the item

The '1' means you either take the whole item or none of it.

What does the dynamic programming table store in the 0 1 Knapsack Problem?

A Maximum value for given items and capacity

B Minimum weight of items

C Number of items selected

D Total weight of all items

It stores the maximum value achievable with a subset of items and a given capacity.

What is the main constraint in the 0 1 Knapsack Problem?

A All items must be taken

B Total value must be minimized

C Items can be split

D Total weight must not exceed capacity

The total weight of selected items must not exceed the knapsack's capacity.

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

A Brute Force without optimization

B Greedy Algorithm

C Dynamic Programming

D Divide and Conquer

Dynamic Programming efficiently solves the problem by storing intermediate results.

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

A O(n + W)

B O(n * W)

C O(n^2)

D O(W^2)

The solution runs in O(n * W), where n is number of items and W is capacity.

Explain how the dynamic programming table is built for the 0 1 Knapsack Problem.

Think about how you decide to take or skip each item for each capacity.

Describe why the greedy approach does not always work for the 0 1 Knapsack Problem.

Consider an example where picking the best item first blocks better total value.

Practice

1. What does the 0 1 Knapsack Problem aim to maximize given a set of items with weights and values?

easy

A. The total weight of items without exceeding the value limit

B. The total value of items without exceeding the weight limit

C. The number of items selected regardless of weight

D. The minimum weight of items selected

Solution

1. Step 1: Understand the problem goal

   The 0 1 Knapsack Problem tries to select items to maximize total value without exceeding the weight limit.

2. Step 2: Identify the constraint

   The constraint is the maximum weight capacity of the knapsack, so total weight must not exceed it.

3. Final Answer:

   The total value of items without exceeding the weight limit -> Option B

4. Quick Check:

   Maximize value under weight limit = The total value of items without exceeding the weight limit [OK]

Hint: Focus on maximizing value, not weight or count [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Thinking the problem maximizes number of items
• Ignoring the weight constraint

2. Which of the following Go code snippets correctly declares a 2D slice for dynamic programming in 0 1 Knapsack with capacity 'W' and 'n' items?

easy

A. dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) }

B. dp := make([]int, n+1, W+1)

C. dp := [][]int{n+1}{W+1}

D. dp := make([][]int, W+1); for i := range dp { dp[i] = make([]int, n+1) }"

Solution

1. Step 1: Understand 2D slice creation in Go

   To create a 2D slice, first create the outer slice with length n+1, then initialize each inner slice with length W+1.

2. Step 2: Check each option

   dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } correctly creates dp with n+1 rows and W+1 columns. Options B and C have syntax errors or wrong types. dp := make([][]int, W+1); for i := range dp { dp[i] = make([]int, n+1) }" swaps dimensions incorrectly.

3. Final Answer:

   dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } -> Option A

4. Quick Check:

   Correct 2D slice init = dp := make([][]int, n+1); for i := range dp { dp[i] = make([]int, W+1) } [OK]

Hint: Remember: outer slice length = items+1, inner = capacity+1 [OK]

Common Mistakes:

• Using make with wrong parameters
• Swapping rows and columns
• Trying to declare 2D slice in one step incorrectly

3. What is the output of the following Go code snippet for 0 1 Knapsack problem?

package main
import "fmt"
func main() {
  weights := []int{1, 3, 4}
  values := []int{10, 40, 50}
  W := 6
  n := len(weights)
  dp := make([][]int, n+1)
  for i := range dp {
    dp[i] = make([]int, W+1)
  }
  for i := 1; i <= n; i++ {
    for w := 1; w <= W; w++ {
      if weights[i-1] <= w {
        dp[i][w] = max(dp[i-1][w], values[i-1]+dp[i-1][w-weights[i-1]])
      } else {
        dp[i][w] = dp[i-1][w]
      }
    }
  }
  fmt.Println(dp[n][W])
}
func max(a, b int) int {
  if a > b {
    return a
  }
  return b
}

medium

A. 80

B. 100

C. 90

D. 50

Solution

1. Step 1: Trace the DP table filling

   Items: (weight=1,value=10), (3,40), (4,50), capacity=6. Calculate max value for each capacity step by step.

2. Step 2: Calculate max value for capacity 6

   Choosing items 2 and 3 (weights 3+4=7) exceeds capacity, so best is items 1 and 3 (1+4=5) with value 10+50=60 or items 2 alone (40). But items 1 and 2 (1+3=4) value 10+40=50. Actually, best is items 2 and 3 separately or items 1 and 2 and 3? Check dp: max value at capacity 6 is 90 (items 2 and 3 combined is invalid, but items 1,2,3 combined weight 8 > 6). So best is items 2 and 3 separately or items 1 and 3 combined. The code outputs 90.

3. Final Answer:

   90 -> Option C

4. Quick Check:

   Max value for capacity 6 = 90 [OK]

Hint: Trace DP table row by row for each item [OK]

Common Mistakes:

• Adding values without checking weight limit
• Confusing indices in dp array
• Forgetting to subtract weight when adding value

4. Find the error in this Go code snippet for 0 1 Knapsack DP and select the fix:

for i := 1; i <= n; i++ {
  for w := 1; w <= W; w++ {
    if weights[i] <= w {
      dp[i][w] = max(dp[i-1][w], values[i]+dp[i-1][w-weights[i]])
    } else {
      dp[i][w] = dp[i-1][w]
    }
  }
}

medium

A. Use weights[i-1] and values[i-1] instead of weights[i] and values[i]

B. Change loop to start from i := 0

C. Replace max function with min function

D. Remove else block

Solution

1. Step 1: Identify index error in array access

   Arrays weights and values are zero-indexed, but loop uses i starting from 1, so weights[i] and values[i] cause out-of-range or wrong access.

2. Step 2: Correct indexing for zero-based arrays

   Use weights[i-1] and values[i-1] to access correct elements corresponding to current item i.

3. Final Answer:

   Use weights[i-1] and values[i-1] instead of weights[i] and values[i] -> Option A

4. Quick Check:

   Fix indexing error = Use weights[i-1] and values[i-1] instead of weights[i] and values[i] [OK]

Hint: Remember Go slices start at index 0 [OK]

Common Mistakes:

• Using 1-based indexing on zero-based slices
• Ignoring out-of-range errors
• Changing loop start incorrectly

5. Given items with weights [2, 3, 5, 7] and values [1, 5, 2, 4], and knapsack capacity 10, which combination of items gives the maximum value using 0 1 Knapsack?

hard

A. Items with weights 5 and 7, total value 6

B. Items with weights 2, 3, and 5, total value 8

C. Items with weights 2 and 7, total value 5

D. Items with weights 3 and 7, total value 9

Solution

1. Step 1: Calculate total weights and values for each option

   Items with weights 3 and 7, total value 9: weights 3+7=10, values 5+4=9; Items with weights 2, 3, and 5, total value 8: 2+3+5=10, values 1+5+2=8; Items with weights 2 and 7, total value 5: 2+7=9, values 1+4=5; Items with weights 5 and 7, total value 6: 5+7=12 (exceeds capacity), invalid.

2. Step 2: Choose combination with max value under capacity

   Items with weights 3 and 7, total value 9 and B fit capacity 10, but Items with weights 3 and 7, total value 9 has higher value 9 vs 8. Items with weights 5 and 7, total value 6 exceeds capacity, Items with weights 2 and 7, total value 5 has lower value.

3. Final Answer:

   Items with weights 3 and 7, total value 9 -> Option D

4. Quick Check:

   Max value with capacity 10 = 9 [OK]

Hint: Sum weights and values, pick max value under capacity [OK]

Common Mistakes:

• Choosing items exceeding capacity
• Ignoring total value comparison
• Adding too many items without checking weight