0 1 Knapsack Problem in DSA Go

Mental Model

We want to pick items with values and weights to maximize value without exceeding weight limit. Each item can be chosen only once.

Analogy: Imagine packing a backpack for a trip. You have limited space and many items. You want to pack the most valuable things without overloading the bag.

Items: [weight: 2, value: 3], [weight: 3, value: 4], [weight: 4, value: 5]
Knapsack capacity: 5

Backpack capacity slots: 0 1 2 3 4 5
Values stored for each capacity as we decide items ->

Dry Run Walkthrough

Input: items: [(weight=2, value=3), (weight=3, value=4), (weight=4, value=5)], capacity=5

Goal: Find the maximum value we can carry without exceeding capacity 5, choosing each item at most once.

Step 1: Start with no items, all capacities have value 0

Capacity: 0 1 2 3 4 5
Value:    0 0 0 0 0 0

Why: We initialize the table with zero value when no items are chosen.

Step 2: Consider first item (weight=2, value=3), update capacities ≥ 2

Capacity: 0 1 2 3 4 5
Value:    0 0 3 3 3 3

Why: For capacity 2 and above, we can take item 1 and get value 3.

Step 3: Consider second item (weight=3, value=4), update capacities ≥ 3

Capacity: 0 1 2 3 4 5
Value:    0 0 3 4 4 7

Why: At capacity 3, choose item 2 alone (value 4). At capacity 5, choose items 1 and 2 (3+4=7).

Step 4: Consider third item (weight=4, value=5), update capacities ≥ 4

Capacity: 0 1 2 3 4 5
Value:    0 0 3 4 5 7

Why: At capacity 4, item 3 alone (value 5) is better than previous 4. At capacity 5, best remains 7 from items 1 and 2.

Result:

Capacity: 0 1 2 3 4 5
Value:    0 0 3 4 5 7
Maximum value = 7

Annotated Code

DSA Go

package main

import "fmt"

// Item holds weight and value
type Item struct {
	weight int
	value  int
}

// knapsack01 returns max value for given items and capacity
func knapsack01(items []Item, capacity int) int {
	// dp array holds max value for each capacity
	dp := make([]int, capacity+1)

	// for each item
	for _, item := range items {
		// traverse capacities backward to avoid reuse
		for w := capacity; w >= item.weight; w-- {
			// check if taking item improves value
			if dp[w-item.weight]+item.value > dp[w] {
				dp[w] = dp[w-item.weight] + item.value
			}
		}
	}
	return dp[capacity]
}

func main() {
	items := []Item{{2, 3}, {3, 4}, {4, 5}}
	capacity := 5
	maxVal := knapsack01(items, capacity)
	fmt.Printf("Maximum value = %d\n", maxVal)
}

for _, item := range items {

iterate over each item to consider including it

for w := capacity; w >= item.weight; w-- {

iterate capacities backward to avoid reusing the same item multiple times

if dp[w-item.weight]+item.value > dp[w] {

check if including current item improves max value at capacity w

dp[w] = dp[w-item.weight] + item.value

update dp with new max value including current item

Output

Maximum value = 7

Complexity Analysis

Time: O(n * capacity) because we process each of the n items for each capacity up to the limit

Space: O(capacity) because we store max values for each capacity in a single array

vs Alternative: Compared to a naive recursive approach that tries all subsets (exponential time), this dynamic programming solution is efficient and runs in polynomial time.

Edge Cases

empty items list

returns 0 because no items to choose

DSA Go

for _, item := range items {

capacity zero

returns 0 because no capacity to hold items

DSA Go

dp := make([]int, capacity+1)

items heavier than capacity

items are skipped, max value remains 0

DSA Go

for w := capacity; w >= item.weight; w-- {

When to Use This Pattern

When you need to select items with weights and values to maximize value without exceeding capacity, and each item can be chosen once, reach for the 0 1 Knapsack pattern using dynamic programming.

Common Mistakes

Mistake: Iterating capacities forward instead of backward causing items to be counted multiple times
Fix: Iterate capacities backward from capacity down to item weight to ensure each item is counted once

Summary

Find the maximum total value of items fitting into a limited capacity bag, choosing each item at most once.

Use when you must optimize selection of items with weights and values under a capacity limit without repetition.

The key is to build solutions for smaller capacities and items, updating from the largest capacity backward to avoid reuse.

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