Backtracking Concept and Decision Tree Visualization in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start backtracking by calling the helper function.

DSA Java

public void solve() {
    backtrack([1]);
}

private void backtrack(List<Integer> path) {
    // backtracking logic here
}

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

Anew ArrayList<>()

Bnew LinkedList<>()

Cpath

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Passing null causes NullPointerException.

Passing an existing path may cause incorrect results.

We start backtracking with an empty list to build the solution path step-by-step.

2fill in blank

medium

Complete the code to check if the current path is a solution.

DSA Java

private void backtrack(List<Integer> path) {
    if (path.size() == [1]) {
        System.out.println(path);
        return;
    }
    // continue backtracking
}

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

A0

Bpath.size()

CtargetSize

Dpath.get(0)

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing path size to 0 always fails.

Using path.get(0) causes errors.

We check if the path length equals the target size to decide if it's a complete solution.

3fill in blank

hard

Fix the error in the loop to try all candidates for the next step.

DSA Java

for (int i = 0; i < [1]; i++) {
    if (!path.contains(candidates[i])) {
        path.add(candidates[i]);
        backtrack(path);
        path.remove(path.size() - 1);
    }
}

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

Acandidates.length

Bpath.length

Ccandidates.size()

Dpath.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using path.size() limits the loop incorrectly.

Using candidates.size() causes compilation error.

We loop over all candidates using candidates.length to cover all options.

4fill in blank

hard

Fill both blanks to add and remove the last element in backtracking.

DSA Java

path.[1](candidates[i]);
backtrack(path);
path.[2](path.size() - 1);

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

Aadd

Bremove

Cappend

Ddelete

Attempts:

3 left

💡 Hint

Common Mistakes

Using append or delete causes compilation errors.

Removing by value instead of index breaks backtracking.

We add the candidate to path, recurse, then remove it to backtrack.

5fill in blank

hard

Fill all three blanks to build a decision tree node with children.

DSA Java

class TreeNode {
    int val;
    List<TreeNode> [1] = new ArrayList<>();

    TreeNode(int val) {
        this.val = [2];
    }

    void addChild(TreeNode child) {
        [3].add(child);
    }
}

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

Achildren

Bval

DchildNodes

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variable names causes errors.

Forgetting to initialize the children list.

The list is named children, val is assigned, and children list adds child nodes.

Practice

1. What is the main idea behind backtracking in algorithms?

easy

A. Try all possible options and undo choices when they lead to dead ends

B. Sort the input data before processing

C. Use a queue to process elements in order

D. Divide the problem into smaller independent parts

Solution

1. Step 1: Understand backtracking approach

   Backtracking tries all possible choices step-by-step and goes back if a choice fails.

2. Step 2: Compare options with backtracking definition

   Only Try all possible options and undo choices when they lead to dead ends describes trying options and undoing wrong choices, which is backtracking.

3. Final Answer:

   Try all possible options and undo choices when they lead to dead ends -> Option A

4. Quick Check:

   Backtracking = Try and undo choices [OK]

Hint: Backtracking means try, fail, and undo to try again [OK]

Common Mistakes:

• Confusing backtracking with sorting or dividing problems
• Thinking backtracking uses queues or simple loops
• Believing backtracking never undoes choices

2. Which of the following is the correct syntax to start a backtracking function in Java?

easy

A. backtrack() void { ... }

B. void backtrack(int step) { ... }

C. backtrack(int step) { ... }

D. function backtrack(int step) { ... }

Solution

1. Step 1: Recall Java method syntax

   Java methods start with return type, then name, then parameters in parentheses.

2. Step 2: Check each option

   void backtrack(int step) { ... } uses correct Java syntax: return type void, method name, parameter list. A: backtrack() void { ... } has wrong order, B: backtrack(int step) { ... } misses return type, D uses 'function' keyword which is not Java.

3. Final Answer:

   void backtrack(int step) { ... } -> Option B

4. Quick Check:

   Java method syntax = void backtrack(int step) { ... } [OK]

Hint: Java methods start with return type, then name, then parameters [OK]

Common Mistakes:

• Using 'function' keyword like JavaScript
• Placing return type after method name
• Omitting the return type

3. What will be the output of the following Java backtracking code snippet?

public class Test {
  static void backtrack(int step) {
    if (step == 3) {
      System.out.print(step + " ");
      return;
    }
    System.out.print(step + " ");
    backtrack(step + 1);
    System.out.print(step + " ");
  }
  public static void main(String[] args) {
    backtrack(1);
  }
}

medium

A. 1 2 1

B. 1 2 3

C. 3 2 1

D. 1 2 3 2 1

Solution

1. Step 1: Trace forward calls and prints

   backtrack(1) prints '1 ', calls backtrack(2). backtrack(2) prints '2 ', calls backtrack(3). backtrack(3) prints '3 ' and returns.

2. Step 2: Trace backward prints after recursion

   After backtrack(3) returns, backtrack(2) prints '2 ', then backtrack(1) prints '1 ' after its recursive call.

3. Final Answer:

   1 2 3 2 1 -> Option D

4. Quick Check:

   Print before and after recursion = 1 2 3 2 1 [OK]

Hint: Print before and after recursion shows mirrored output [OK]

Common Mistakes:

• Ignoring prints after recursive call
• Assuming only forward prints happen
• Confusing base case output

4. Identify the error in this backtracking code snippet:

void backtrack(int step) {
  if (step > 3) return;
  System.out.print(step + " ");
  backtrack(step + 1);
}

medium

A. Missing base case to stop recursion

B. No print statement after recursion to backtrack

C. No error, code works correctly

D. Base case condition is incorrect, should be step == 3

Solution

1. Step 1: Analyze base case condition

   The base case stops recursion when step > 3, so calls happen for step=1,2,3 then stop.

2. Step 2: Check print and recursion flow

   Print happens before recursion, so output is '1 2 3 '. No error in stopping or printing.

3. Final Answer:

   No error, code works correctly -> Option C

4. Quick Check:

   Base case and print correct = No error [OK]

Hint: Check base case carefully to confirm recursion stops [OK]

Common Mistakes:

• Thinking base case must be step == 3
• Expecting print after recursion for backtracking
• Assuming missing braces cause error

5. Given a decision tree for backtracking with choices A and B at each step up to 2 steps, how many total paths (leaf nodes) will the tree have?

medium

A. 4

B. 3

C. 2

D. 8

Solution

1. Step 1: Understand decision tree branching

   Each step has 2 choices (A or B), so the tree branches twice per step.

2. Step 2: Calculate total leaf nodes for 2 steps

   Number of leaf nodes = 2 choices ^ 2 steps = 2 * 2 = 4 total paths.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   2 choices ^ 2 steps = 4 [OK]

Hint: Total paths = choices^steps in decision tree [OK]

Common Mistakes:

• Adding instead of multiplying choices
• Confusing steps with choices
• Forgetting to count all leaf nodes