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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

function backtrack(path: number[], nums: number[]): void {
  // base case and recursion logic here
}

function solve(nums: number[]): void {
  backtrack([1], nums);
}

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

A[1]

B[]

C[nums]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Starting with a non-empty path like [0] or [1] which is incorrect.

Passing the whole nums array as path.

We start backtracking with an empty path, so we pass an empty array [].

2fill in blank

medium

Complete the code to add the current number to the path inside backtracking.

DSA Typescript

function backtrack(path: number[], nums: number[]): void {
  for (let i = 0; i < nums.length; i++) {
    path.push(nums[i]);
    // recursive call
    backtrack(path, nums);
    path.[1]();
  }
}

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

Apop

Bpush

Cshift

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using push() instead of pop() which adds instead of removes.

Using shift() or unshift() which remove or add elements at the start.

After exploring with the current number, we remove it using pop() to backtrack.

3fill in blank

hard

Fix the error in the base case condition to stop recursion when path length equals nums length.

DSA Typescript

function backtrack(path: number[], nums: number[]): void {
  if (path.length [1] nums.length) {
    console.log(path);
    return;
  }
  // recursion continues
}

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

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or > which do not correctly detect completion.

Using != which triggers at wrong times.

We stop when path length equals nums length, so the condition is path.length == nums.length.

4fill in blank

hard

Fill both blanks to create a decision tree node with current path and index.

DSA Typescript

interface TreeNode {
  path: number[];
  index: number;
}

function createNode(path: number[], idx: number): TreeNode {
  return {
    path: path[1],
    index: [2]
  };
}

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

A.slice()

B.push()

Cidx

Dpath.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using push() which modifies the original array.

Using path.length instead of idx for index.

We copy the path using slice() to avoid mutation, and set index to idx parameter.

5fill in blank

hard

Fill all three blanks to build a map of paths to their next choices in the decision tree.

DSA Typescript

function buildDecisionTree(nums: number[]): Map<string, number[]> {
  const tree = new Map<string, number[]>();
  function backtrack(path: number[]): void {
    const key = path.[1]();
    tree.set(key, []);
    for (let num of nums) {
      if (!path.includes(num)) {
        tree.get(key)!.[2](num);
        backtrack(path.[3](num));
      }
    }
  }
  backtrack([]);
  return tree;
}

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

Ajoin(',')

Bpush

Cconcat

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which removes elements instead of adding.

Using push() on path array instead of map array.

Using join without separator causing ambiguous keys.

We join path to string key, push next numbers to map array, and concat num to path for recursion.

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 data before processing it

C. Use loops to repeat actions without conditions

D. Store data in a fixed-size array

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, undo, and try again [OK]

Common Mistakes:

• Confusing backtracking with sorting
• Thinking backtracking is just looping
• Believing backtracking stores fixed data only

2. Which of the following is the correct TypeScript function signature for a backtracking function that tries to build a string path?

easy

A. function backtrack(path: number): string

B. function backtrack(path: string): void

C. function backtrack(): number[]

D. function backtrack(path: boolean): boolean

Solution

1. Step 1: Identify expected input and output

   Backtracking often builds a path string and does not return a value, so void return is common.

2. Step 2: Match signature with backtracking pattern

   function backtrack(path: string): void matches a function taking a string path and returning void, which fits backtracking.

3. Final Answer:

   function backtrack(path: string): void -> Option B

4. Quick Check:

   Backtracking function signature = string input, void output [OK]

Hint: Backtracking functions usually return void and take current path [OK]

Common Mistakes:

• Using wrong return types like number or boolean
• Missing input parameter for current path
• Confusing return type with output data

3. Consider this TypeScript backtracking snippet:

function backtrack(path: string, choices: string[]): void {
  if (path.length === 2) {
    console.log(path);
    return;
  }
  for (const choice of choices) {
    backtrack(path + choice, choices);
  }
}
backtrack('', ['A', 'B']);

What will be printed?

medium

A. AA, AB, BA, BB

B. A, B

C. AB, BA

D. AA, BB

Solution

1. Step 1: Understand recursion and stopping condition

   The function prints when path length is 2, building strings by adding 'A' or 'B'.

2. Step 2: Trace all paths of length 2

   Starting empty, choices add 'A' or 'B' twice: 'AA', 'AB', 'BA', 'BB' are printed.

3. Final Answer:

   AA, AB, BA, BB -> Option A

4. Quick Check:

   All 2-length combos of A and B printed [OK]

Hint: Count all 2-letter combos from choices [OK]

Common Mistakes:

• Printing only single letters
• Missing some combinations
• Confusing order of printed strings

4. This backtracking code tries to generate all subsets of [1, 2] but has a bug:

function backtrack(start: number, path: number[], nums: number[]) {
  console.log(path);
  for (let i = start; i < nums.length; i++) {
    path.push(nums[i]);
    backtrack(i, path, nums);
    path.pop();
  }
}
backtrack(0, [], [1, 2]);

What is the bug?

medium

A. start should be initialized to 1

B. path.pop() is missing to undo choice

C. Console.log should be after the loop

D. Recursive call should use i + 1 instead of i to avoid duplicates

Solution

1. Step 1: Analyze recursive call parameter

   The call uses backtrack(i, ...) which reuses the same index, causing repeated subsets.

2. Step 2: Fix recursion to avoid duplicates

   Changing to backtrack(i + 1, ...) moves forward, preventing duplicates and infinite loops.

3. Final Answer:

   Recursive call should use i + 1 instead of i to avoid duplicates -> Option D

4. Quick Check:

   Advance index in recursion = i + 1 [OK]

Hint: Use i + 1 in recursion to move forward [OK]

Common Mistakes:

• Not advancing index causing infinite recursion
• Removing path.pop() causing wrong subsets
• Misplacing console.log outside recursion

5. You want to use backtracking to find all 3-letter strings from letters ['X', 'Y', 'Z'] but without repeating the same letter consecutively. Which change to this code snippet enforces that?

function backtrack(path: string, letters: string[]) {
  if (path.length === 3) {
    console.log(path);
    return;
  }
  for (const letter of letters) {
    // What condition to add here?
    backtrack(path + letter, letters);
  }
}
backtrack('', ['X', 'Y', 'Z']);

hard

A. Add: if (path.length < 3) before recursive call

B. Add: if (letter !== 'X') before recursive call

C. Add: if (path.length === 0 || path[path.length - 1] !== letter) before recursive call

D. Add: if (path.includes(letter)) before recursive call

Solution

1. Step 1: Understand the no consecutive repeat rule

   We must skip adding a letter if it matches the last letter in path.

2. Step 2: Write condition to check last letter

   Condition if path is empty or last letter differs from current letter ensures no repeats.

3. Final Answer:

   Add: if (path.length === 0 || path[path.length - 1] !== letter) before recursive call -> Option C

4. Quick Check:

   Check last letter before adding new letter [OK]

Hint: Check last letter to avoid consecutive repeats [OK]

Common Mistakes:

• Filtering letters incorrectly by fixed letter
• Checking only path length without letter comparison
• Using includes() which disallows all repeats