Why Greedy Works and When It Fails in DSA Go - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the next item with the highest value per weight ratio in a greedy approach.

DSA Go

func selectNextItem(items []Item) Item {
    var best Item
    bestRatio := 0.0
    for _, item := range items {
        ratio := float64(item.Value) / float64(item.Weight)
        if ratio > [1] {
            best = item
            bestRatio = ratio
        }
    }
    return best
}

Drag options to blanks, or click blank then click option'

Aitem.Weight

BbestRatio

Citem.Value

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing ratio to 0 instead of bestRatio.

Using item.Value or item.Weight directly in the if condition.

We compare the current item's value-to-weight ratio with the bestRatio found so far to pick the best item greedily.

2fill in blank

medium

Complete the code to check if the greedy solution is optimal by comparing total weight with capacity.

DSA Go

func canAddItem(currentWeight, itemWeight, capacity int) bool {
    return currentWeight + itemWeight [1] capacity
}

Drag options to blanks, or click blank then click option'

A<=

B<

C>=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= causes off-by-one errors.

Using > or >= reverses the logic.

We allow adding the item if total weight does not exceed capacity, so use <=.

3fill in blank

hard

Fix the error in the greedy coin change function to correctly pick the largest coin not exceeding the amount.

DSA Go

func greedyCoinChange(coins []int, amount int) []int {
    var result []int
    for _, coin := range coins {
        for amount [1] coin {
            amount -= coin
            result = append(result, coin)
        }
    }
    return result
}

Drag options to blanks, or click blank then click option'

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of >= causes the loop to never run.

Using > excludes the case when amount equals coin.

We subtract coin from amount while amount is greater than or equal to coin value.

4fill in blank

hard

Fill both blanks to create a map comprehension that filters words longer than 3 characters and maps them to their lengths.

DSA Go

lengths := map[string]int{}
for _, word := range words {
    if len(word) [1] 3 {
        lengths[word] = [2]
    }
}

Drag options to blanks, or click blank then click option'

A>

Blen(word)

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using == or < instead of > in the condition.

Assigning word instead of len(word) in the map.

We check if word length is greater than 3 and map word to its length.

5fill in blank

hard

Fill all three blanks to create a map of uppercase words to their lengths, filtering words longer than 4 characters.

DSA Go

result := map[string]int{}
for _, word := range words {
    if len(word) [1] 4 {
        result[[2]] = [3]
    }
}

Drag options to blanks, or click blank then click option'

A>

Blen(word)

Cstrings.ToUpper(word)

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= instead of > in the condition.

Assigning word instead of uppercase word as key.

Assigning word instead of len(word) as value.

We filter words longer than 4, map uppercase word to its length.

Practice

1. What is the main reason a greedy algorithm works correctly for some problems?

easy

A. Because it sorts the input data before processing

B. Because it tries all possible solutions and picks the best one

C. Because it makes the best choice at each step hoping for a global optimum

D. Because it uses recursion to explore all paths

Solution

1. Step 1: Understand greedy choice property

   A greedy algorithm makes the best local choice at each step, aiming for a global best solution.

2. Step 2: Differentiate from exhaustive search

   Unlike trying all solutions or recursion, greedy does not explore all paths but relies on local optimality.

3. Final Answer:

   Because it makes the best choice at each step hoping for a global optimum -> Option C

4. Quick Check:

   Greedy choice property = Because it makes the best choice at each step hoping for a global optimum [OK]

Hint: Greedy picks best local choice, not all possibilities [OK]

Common Mistakes:

• Confusing greedy with exhaustive search
• Thinking greedy always tries all solutions
• Assuming sorting alone makes greedy correct

2. Which of the following Go code snippets correctly implements a greedy step to pick the largest number from a slice?

easy

A. max := nums[0]
for _, v := range nums {
if v > max {
max = v
}
}

B. max := 0
for i := 1; i < len(nums); i++ {
if nums[i] < max {
max = nums[i]
}
}

C. max := nums[0]
for i := 0; i < len(nums); i++ {
if nums[i] < max {
max = nums[i]
}
}

D. max := nums[0]
for _, v := range nums {
if v < max {
max = v
}
}

Solution

1. Step 1: Identify correct max initialization and comparison

   Initialize max with first element and update if current value is greater.

2. Step 2: Check loop and condition correctness

   max := nums[0] for _, v := range nums { if v > max { max = v } } uses correct range loop and updates max when v > max.

3. Final Answer:

   Option A correctly finds the largest number -> Option A

4. Quick Check:

   Correct max update condition = max := nums[0] for _, v := range nums { if v > max { max = v } } [OK]

Hint: Update max only if current value is greater [OK]

Common Mistakes:

• Using '<' instead of '>' to find max
• Initializing max to zero instead of first element
• Comparing incorrectly inside loop

3. What will be the output of the following Go code that tries to select intervals greedily by earliest finish time?

package main
import "fmt"
func main() {
  intervals := [][]int{{1,4},{3,5},{0,6},{5,7},{8,9},{5,9}}
  selected := [][]int{}
  lastEnd := 0
  for _, interval := range intervals {
    if interval[0] >= lastEnd {
      selected = append(selected, interval)
      lastEnd = interval[1]
    }
  }
  fmt.Println(selected)
}

medium

A. [[1 4] [5 7] [8 9]]

B. [[1 4] [3 5] [5 7] [8 9]]

C. [[0 6] [8 9]]

D. [[1 4] [5 9]]

Solution

1. Step 1: Understand greedy selection by earliest finish time

   Intervals are checked in given order; select if start >= last selected end.

2. Step 2: Trace selection step-by-step

   Selected intervals: [1,4] (lastEnd=4), skip [3,5], skip [0,6], select [5,7] (lastEnd=7), select [8,9].

3. Final Answer:

   [[1 4] [5 7] [8 9]] -> Option A

4. Quick Check:

   Greedy interval selection = [[1 4] [5 7] [8 9]] [OK]

Hint: Pick intervals starting after last end time [OK]

Common Mistakes:

• Not updating lastEnd correctly
• Selecting overlapping intervals
• Assuming input is sorted by finish time

4. The following Go code tries to solve the coin change problem greedily but sometimes fails. What is the bug?

package main
import "fmt"
func main() {
  coins := []int{25, 10, 5, 1}
  amount := 30
  count := 0
  for _, coin := range coins {
    count += amount / coin
    amount = amount % coin
  }
  fmt.Println(count)
}

medium

A. The coins slice is not sorted in ascending order

B. The greedy approach fails for some coin sets, but code logic is correct

C. The code does not update amount correctly inside the loop

D. The count variable should be reset inside the loop

Solution

1. Step 1: Analyze greedy coin change logic

   The code picks largest coins first, counting how many fit, updating amount correctly.

2. Step 2: Understand greedy failure cases

   Greedy works for US coins but fails for some coin sets; code logic is correct but approach can fail.

3. Final Answer:

   The greedy approach fails for some coin sets, but code logic is correct -> Option B

4. Quick Check:

   Greedy correctness depends on coin set = The greedy approach fails for some coin sets, but code logic is correct [OK]

Hint: Greedy coin change fails on non-canonical coin sets [OK]

Common Mistakes:

• Thinking code has syntax or logic error
• Assuming coin order must be ascending
• Resetting count inside loop incorrectly

5. You want to schedule the maximum number of non-overlapping meetings from a list of start and end times. Which greedy strategy guarantees the optimal solution?

hard

A. Always pick the meeting with the latest start time

B. Always pick the meeting with the shortest duration

C. Always pick the meeting with the earliest start time

D. Always pick the meeting with the earliest finish time

Solution

1. Step 1: Recall interval scheduling greedy strategy

   The proven greedy method is to pick meetings finishing earliest to maximize remaining time.

2. Step 2: Compare other strategies

   Picking earliest start or shortest duration can cause overlaps and miss optimal count.

3. Final Answer:

   Always pick the meeting with the earliest finish time -> Option D

4. Quick Check:

   Earliest finish time strategy = Always pick the meeting with the earliest finish time [OK]

Hint: Pick meetings finishing earliest to maximize schedule [OK]

Common Mistakes:

• Choosing earliest start time instead of finish time
• Picking shortest duration meetings incorrectly
• Assuming latest start time works best