Why Backtracking and What Greedy Cannot Solve in DSA Go - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the backtracking recursion with the initial empty solution.

DSA Go

func backtrack(solution []int, choices []int) {
    if len(solution) == len(choices) {
        fmt.Println(solution)
        return
    }
    for _, choice := range choices {
        if !contains(solution, choice) {
            backtrack(append(solution, choice), choices)
        }
    }
}

func main() {
    choices := []int{1, 2, 3}
    backtrack([1], choices)
}

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

Aappend(solution, choice)

B[]int{}

Cchoices

Dsolution

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the full choices slice as the initial solution.

Passing nil instead of an empty slice.

The backtracking starts with an empty solution slice, so we pass an empty slice []int{} to the backtrack function.

2fill in blank

medium

Complete the code to check if a choice is already in the current solution.

DSA Go

func contains(slice []int, val int) bool {
    for _, item := range slice {
        if item [1] val {
            return true
        }
    }
    return false
}

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using != which checks for inequality.

Using < or > which are not correct for equality check.

We check if the current item is equal to the value to find if it is already in the solution.

3fill in blank

hard

Fix the error in the backtracking function call to correctly add the choice to the solution.

DSA Go

func backtrack(solution []int, choices []int) {
    if len(solution) == len(choices) {
        fmt.Println(solution)
        return
    }
    for _, choice := range choices {
        if !contains(solution, choice) {
            backtrack(append(solution[1]choice), choices)
        }
    }
}

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

A,

Bappend(solution, choice)

C+

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using + which is not valid for slices.

Using append without parentheses.

In Go, to add an element to a slice in a function call, we use append(solution, choice). The blank should be a comma to separate the function arguments inside append.

4fill in blank

hard

Fill both blanks to create a map that filters numbers greater than 3 and squares them.

DSA Go

result := map[int]int{}
for _, num := range []int{1, 2, 3, 4, 5} {
    if num [1] 3 {
        result[num] = num [2] 2
    }
}

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

A>

B*

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > in the condition.

Using + instead of * for squaring.

The condition filters numbers greater than 3 using >. The square is calculated by multiplying the number by itself using *.

5fill in blank

hard

Fill all three blanks to create a map of uppercase keys to values filtered by value > 0.

DSA Go

data := map[string]int{"a": 1, "b": -1, "c": 3}
result := map[string]int{}
for k, v := range data {
    if v [3] 0 {
        result[[1]] = [2]
    }
}

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

Astrings.ToUpper(k)

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > in the condition.

Using k instead of strings.ToUpper(k) for keys.

Assigning k instead of v as value.

The condition filters values greater than 0 using >. The key is converted to uppercase using strings.ToUpper(k). The value is assigned as is with v.

Practice

1. Why do we need backtracking instead of just using a greedy approach for some problems?

easy

A. Because backtracking only works for sorting problems.

B. Because backtracking always runs faster than greedy algorithms.

C. Because greedy algorithms use more memory than backtracking.

D. Because greedy algorithms make local choices that may not lead to the best overall solution.

Solution

1. Step 1: Understand greedy algorithm behavior

   Greedy algorithms pick the best option at each step without reconsidering previous choices.

2. Step 2: Understand backtracking advantage

   Backtracking tries all possibilities and can undo choices to find the best overall solution.

3. Final Answer:

   Because greedy algorithms make local choices that may not lead to the best overall solution. -> Option D

4. Quick Check:

   Greedy local choice ≠ global best [OK]

Hint: Backtracking explores all options; greedy picks locally best only [OK]

Common Mistakes:

• Thinking greedy is always faster and better
• Confusing memory use between algorithms
• Believing backtracking is only for sorting

2. Which of the following Go code snippets correctly starts a backtracking function signature for solving a problem?

easy

A. func backtrack(path []int, choices []int) { }

B. func backtrack(path int, choices int) { }

C. func backtrack() int { }

D. func backtrack(path string) string { }

Solution

1. Step 1: Identify typical backtracking parameters

   Backtracking usually passes the current path (partial solution) and remaining choices as slices.

2. Step 2: Check each function signature

   func backtrack(path []int, choices []int) { } uses slices for path and choices, suitable for backtracking. Others use single values or wrong types.

3. Final Answer:

   func backtrack(path []int, choices []int) { } -> Option A

4. Quick Check:

   Backtracking uses slices for path and choices [OK]

Hint: Backtracking functions usually take slices for path and choices [OK]

Common Mistakes:

• Using single int instead of slices
• Returning values when not needed
• Using wrong parameter types

3. What is the output of this Go code snippet that tries to solve the subset sum problem using backtracking?

package main
import "fmt"

func backtrack(nums []int, target int, start int, path []int) {
    if target == 0 {
        fmt.Println(path)
        return
    }
    for i := start; i < len(nums); i++ {
        if nums[i] > target {
            continue
        }
        backtrack(nums, target-nums[i], i+1, append(path, nums[i]))
    }
}

func main() {
    nums := []int{2, 3, 6, 7}
    backtrack(nums, 7, 0, []int{})
}

medium

A. [2 3 6]
[7]

B. [2 3 7]
[7]

C. [7]

D. [7]
[2 3 6]

Solution

1. Step 1: Understand the problem and code

   The code finds subsets of nums that sum to 7, using backtracking and no repeats.

2. Step 2: Trace the backtracking calls

   Possible subsets: [7] sums to 7; [2,3,6] sums to 11 (too big), so skipped; only [7] prints.

3. Final Answer:

   [7] -> Option C

4. Quick Check:

   Only subset summing to 7 is [7] [OK]

Hint: Check sums carefully; skip subsets exceeding target [OK]

Common Mistakes:

• Assuming [2 3 6] sums to 7
• Confusing order of printed subsets
• Ignoring the continue condition

4. Find the bug in this Go backtracking code snippet that tries to generate all permutations of a slice of integers:

func permute(nums []int, path []int, used []bool) {
    if len(path) == len(nums) {
        fmt.Println(path)
        return
    }
    for i := 0; i < len(nums); i++ {
        if used[i] {
            continue
        }
        path = append(path, nums[i])
        used[i] = true
        permute(nums, path, used)
        used[i] = false
    }
}

medium

A. Missing removal of last element from path after recursion

B. Used slice is not initialized

C. Base case condition is incorrect

D. Loop index should start from 1

Solution

1. Step 1: Analyze backtracking recursion steps

   After recursive call, path must remove last added element to backtrack correctly.

2. Step 2: Identify missing backtracking step

   Code appends to path but never removes last element after recursion, causing incorrect paths.

3. Final Answer:

   Missing removal of last element from path after recursion -> Option A

4. Quick Check:

   Backtracking needs to undo path changes [OK]

Hint: Always undo changes to path after recursive call [OK]

Common Mistakes:

• Forgetting to remove last element from path
• Not initializing used slice before call
• Incorrect base case length check

5. You want to solve the classic N-Queens problem where you place N queens on an NxN chessboard so that no two queens attack each other. Why can't a greedy algorithm solve this correctly, and how does backtracking help?

hard

A. Greedy only works for even N; backtracking works for all N.

B. Greedy places queens row by row without checking future conflicts; backtracking tries all placements and backtracks on conflicts.

C. Greedy uses too much memory; backtracking uses less memory by pruning.

D. Greedy sorts queens by position; backtracking sorts by attack count.

Solution

1. Step 1: Understand greedy limitations for N-Queens

   Greedy tries to place queens row by row choosing first safe spot, but may miss solutions requiring backtracking.

2. Step 2: Understand backtracking approach

   Backtracking tries all possible queen placements, undoing choices when conflicts arise, ensuring all solutions are found.

3. Final Answer:

   Greedy places queens row by row without checking future conflicts; backtracking tries all placements and backtracks on conflicts. -> Option B

4. Quick Check:

   N-Queens needs full search, not greedy local choice [OK]

Hint: N-Queens needs full search; greedy fails on future conflicts [OK]

Common Mistakes:

• Thinking greedy can always find a solution
• Confusing memory use differences
• Believing greedy works only for even N