Backtracking Concept and Decision Tree Visualization in DSA Typescript - Build from Scratch

Backtracking Concept and Decision Tree Visualization

📖 Scenario: Imagine you are helping a friend solve a simple puzzle where they must choose numbers from a list to reach a target sum. You will use backtracking to explore all possible choices step-by-step and visualize the decision tree of choices made.

🎯 Goal: Build a TypeScript program that uses backtracking to find all combinations of numbers that sum up to a target. Visualize the decision tree by printing the current path at each step.

📋 What You'll Learn

Create an array of numbers called candidates with values [2, 3, 6, 7]

Create a variable called target with value 7

Write a backtracking function called backtrack that tries adding numbers to a current path to reach the target sum

Print each valid combination found as a path

Print the current path at each decision step to visualize the decision tree

💡 Why This Matters

🌍 Real World

Backtracking is used in puzzles, games, and solving problems like finding combinations or permutations where you explore choices step-by-step.

💼 Career

Understanding backtracking helps in algorithm design, coding interviews, and solving complex problems in software development.

1

Create the initial data setup

Create an array called candidates with values [2, 3, 6, 7] and a variable called target with value 7.

DSA Typescript

// Your code here

Hint

Use const candidates: number[] = [2, 3, 6, 7]; and const target: number = 7;

2

Add a variable to hold the current path

Create an empty array of numbers called currentPath to store the current combination of numbers during backtracking.

DSA Typescript

const candidates: number[] = [2, 3, 6, 7];
const target: number = 7;
// Your code here

Hint

Use const currentPath: number[] = []; to hold the current numbers chosen.

3

Write the backtracking function

Write a function called backtrack that takes two parameters: start (number) and remainingTarget (number). Inside, if remainingTarget is 0, print the currentPath. Otherwise, loop over candidates starting from start, add the candidate to currentPath, print the current path to visualize the decision tree, recursively call backtrack with updated parameters, then remove the last number from currentPath to backtrack.

DSA Typescript

const candidates: number[] = [2, 3, 6, 7];
const target: number = 7;
const currentPath: number[] = [];
// Your code here

Hint

Use recursion with base case remainingTarget === 0. Print currentPath at each step to show the decision tree.

4

Call the backtracking function and print results

Call the backtrack function with start as 0 and remainingTarget as target to start the search.

DSA Typescript

const candidates: number[] = [2, 3, 6, 7];
const target: number = 7;
const currentPath: number[] = [];

function backtrack(start: number, remainingTarget: number): void {
    if (remainingTarget === 0) {
        console.log('Found combination:', currentPath.join(' -> '));
        return;
    }
    for (let i = start; i  remainingTarget) continue;
        currentPath.push(candidates[i]);
        console.log('Current path:', currentPath.join(' -> '));
        backtrack(i, remainingTarget - candidates[i]);
        currentPath.pop();
    }
}
// Your code here

Hint

Call backtrack(0, target); to start the search and print all paths.

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