Why 0 1 Knapsack Problem in DSA Go?

The Big Idea

What if you could pack your bag perfectly every time without guessing?

The Scenario

Imagine you have a backpack that can carry only a limited weight. You want to pack the most valuable items, but you can either take an item whole or leave it behind. How do you choose which items to pack?

The Problem

Trying every possible combination by hand is slow and confusing. You might miss the best choice or waste time checking useless options. It's easy to make mistakes and hard to keep track of what you tried.

The Solution

The 0 1 Knapsack Problem uses a smart way to check all choices quickly without repeating work. It helps find the best set of items to maximize value while staying within the weight limit.

Before vs After

✗ Before

func knapsack(weights []int, values []int, capacity int) int {
    maxValue := 0
    // Try all subsets manually (too many!)
    // ...
    return maxValue
}

✓ After

func knapsack(weights []int, values []int, capacity int) int {
    dp := make([]int, capacity+1)
    for i := 0; i < len(weights); i++ {
        for w := capacity; w >= weights[i]; w-- {
            if dp[w-weights[i]]+values[i] > dp[w] {
                dp[w] = dp[w-weights[i]] + values[i]
            }
        }
    }
    return dp[capacity]
}

What It Enables

This problem-solving method lets you quickly find the best combination of items to carry, saving time and effort.

Real Life Example

Choosing which groceries to buy with a limited budget and bag space, so you get the most useful and valuable items.

Key Takeaways

Manual checking of all item combinations is slow and error-prone.

0 1 Knapsack uses dynamic programming to solve efficiently.

It helps pick the best items to maximize value within weight limits.

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