Why Backtracking and What Greedy Cannot Solve in DSA Typescript - See It Work

Why Backtracking and What Greedy Cannot Solve

📖 Scenario: Imagine you are organizing a small party and want to invite friends with different preferences. You want to find all possible groups of friends that fit a certain budget. Sometimes, picking the cheapest friends first (greedy) won't find the best group. You need to try different combinations carefully (backtracking).

🎯 Goal: You will create a program that tries all possible groups of friends within a budget using backtracking. You will see why a simple greedy approach cannot always find all valid groups.

📋 What You'll Learn

Create a list of friends with their costs

Set a budget limit

Use backtracking to find all groups of friends within the budget

Print all valid groups

💡 Why This Matters

🌍 Real World

Backtracking helps in planning events, scheduling, and solving puzzles where all combinations matter.

💼 Career

Understanding backtracking is key for roles in software development, especially in algorithm design, optimization, and problem solving.

1

Create the list of friends with their costs

Create a constant array called friends with these exact entries: { name: 'Alice', cost: 20 }, { name: 'Bob', cost: 10 }, { name: 'Charlie', cost: 15 }, { name: 'Diana', cost: 25 }

DSA Typescript

// Your code here

Hint

Use an array of objects with name and cost properties.

2

Set the budget limit

Create a constant called budget and set it to 35

DSA Typescript

const friends = [
  { name: 'Alice', cost: 20 },
  { name: 'Bob', cost: 10 },
  { name: 'Charlie', cost: 15 },
  { name: 'Diana', cost: 25 }
];
// Your code here

Hint

Just create a number variable called budget with value 35.

3

Write the backtracking function to find all groups within budget

Write a function called findGroups that takes index, currentGroup, and currentCost as parameters. Use backtracking to explore all friend combinations starting from index. If currentCost is less than or equal to budget and currentGroup is not empty, add the currentGroup to a global array validGroups. Recursively try adding each friend from index onwards. Do not exceed the budget.

DSA Typescript

const friends = [
  { name: 'Alice', cost: 20 },
  { name: 'Bob', cost: 10 },
  { name: 'Charlie', cost: 15 },
  { name: 'Diana', cost: 25 }
];

const budget = 35;

const validGroups: string[][] = [];

// Your code here

Hint

Use recursion and try adding each friend if budget allows. Save copies of valid groups.

4

Call the function and print all valid groups

Call findGroups starting from index 0, empty array [], and cost 0. Then print the validGroups array.

DSA Typescript

const friends = [
  { name: 'Alice', cost: 20 },
  { name: 'Bob', cost: 10 },
  { name: 'Charlie', cost: 15 },
  { name: 'Diana', cost: 25 }
];

const budget = 35;

const validGroups: string[][] = [];

function findGroups(index: number, currentGroup: string[], currentCost: number): void {
  if (currentCost  0) {
    validGroups.push([...currentGroup]);
  }

  for (let i = index; i < friends.length; i++) {
    const friend = friends[i];
    if (currentCost + friend.cost <= budget) {
      currentGroup.push(friend.name);
      findGroups(i + 1, currentGroup, currentCost + friend.cost);
      currentGroup.pop();
    }
  }
}

// Your code here

Hint

Start the search from 0 with empty group and cost 0. Then print the results.

Practice

1. Why is backtracking preferred over greedy algorithms for solving problems like the N-Queens puzzle?

easy

A. Because greedy algorithms always find the optimal solution faster

B. Because backtracking explores all possible solutions and can undo choices, while greedy makes irreversible choices

C. Because backtracking uses less memory than greedy algorithms

D. Because greedy algorithms are only used for sorting problems

Solution

1. Step 1: Understand backtracking's approach

   Backtracking tries all possible options and can go back (undo) if a choice leads to failure.

2. Step 2: Compare with greedy approach

   Greedy algorithms make the best immediate choice but cannot undo it, so they may miss the correct solution.

3. Final Answer:

   Because backtracking explores all possible solutions and can undo choices, while greedy makes irreversible choices -> Option B

4. Quick Check:

   Backtracking = explores all + undo [OK]

Hint: Backtracking tries all paths; greedy picks one path only [OK]

Common Mistakes:

• Thinking greedy always finds the best solution
• Confusing backtracking with brute force
• Believing backtracking uses less memory

2. Which of the following TypeScript code snippets correctly represents the start of a backtracking function signature?

easy

A. function backtrack(path number[], options number[]): void

B. function backtrack(path: number[], options: number[]) => void

C. function backtrack(path: number[], options: number[]): void

D. function backtrack(path: number[], options: number[]): number

Solution

1. Step 1: Check TypeScript function syntax

   Correct syntax uses colon ':' to specify types and parentheses for parameters.

2. Step 2: Verify return type and arrow usage

   function backtrack(path: number[], options: number[]): void uses correct syntax with void return type; others have syntax errors or wrong return type.

3. Final Answer:

   function backtrack(path: number[], options: number[]): void -> Option C

4. Quick Check:

   Correct TypeScript function syntax = function backtrack(path: number[], options: number[]): void [OK]

Hint: Look for colons and parentheses in function parameters [OK]

Common Mistakes:

• Missing colons between parameter and type
• Using arrow '=>' incorrectly in function signature
• Wrong return type specified

3. Consider this TypeScript backtracking code snippet for generating subsets:

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

const nums = [1, 2];
const result: number[][] = [];
backtrack(0, [], nums, result);
console.log(result);

What is the output printed?

medium

A. [[], [2], [1, 2], [1]]

B. [[1], [2]]

C. [[1, 2]]

D. [[], [1], [1, 2], [2]]

Solution

1. Step 1: Trace backtracking calls with nums = [1, 2]

   Start with empty path [], add it to result. Then add 1, recurse, add [1], add 2, recurse, add [1,2], backtrack, add 2 alone, add [2].

2. Step 2: Collect all subsets generated

   Subsets are [], [1], [1, 2], and [2] in order.

3. Final Answer:

   [[], [1], [1, 2], [2]] -> Option D

4. Quick Check:

   All subsets generated = [[], [1], [1, 2], [2]] [OK]

Hint: Backtracking with push/pop generates all subsets including empty [OK]

Common Mistakes:

• Forgetting to add empty subset
• Not popping after recursion causing wrong subsets
• Mixing order of subsets

4. The following TypeScript backtracking code tries to find all permutations of [1, 2], but it has a bug:

function permute(nums: number[], path: number[], used: boolean[], result: number[][]) {
  if (path.length === nums.length) {
    result.push(path);
    return;
  }
  for (let i = 0; i < nums.length; i++) {
    if (used[i]) continue;
    path.push(nums[i]);
    used[i] = true;
    permute(nums, path, used, result);
    path.pop();
    used[i] = false;
  }
}

const result: number[][] = [];
permute([1, 2], [], [false, false], result);
console.log(result);

What is the bug and how to fix it?

medium

A. Bug: Pushing path directly causes reference issues; fix by pushing a copy of path

B. Bug: used array is not updated correctly; fix by removing used[i] = false

C. Bug: Base case condition is wrong; fix by changing path.length === nums.length to path.length > nums.length

D. Bug: Loop should start from 1 instead of 0; fix by changing for loop start

Solution

1. Step 1: Identify the problem with pushing path

   Pushing path directly adds the same array reference multiple times, causing all entries to be the same final path.

2. Step 2: Fix by pushing a copy of path

   Use result.push([...path]) to push a new array copy each time.

3. Final Answer:

   Bug: Pushing path directly causes reference issues; fix by pushing a copy of path -> Option A

4. Quick Check:

   Push copy of path, not reference [OK]

Hint: Always push a copy of path in backtracking results [OK]

Common Mistakes:

• Pushing path reference instead of copy
• Incorrect base case condition
• Misusing used array updates

5. You want to solve the classic "Subset Sum" problem: given an array of positive integers and a target sum, find if any subset sums exactly to the target. Why can't a greedy algorithm solve this correctly in all cases, and how does backtracking help?

hard

A. Greedy fails because picking largest numbers first may miss exact sum; backtracking tries all subsets to find exact match

B. Greedy fails because it sorts the array; backtracking sorts and tries subsets

C. Greedy fails because it uses recursion; backtracking uses iteration

D. Greedy fails because it only works on sorted arrays; backtracking only works on unsorted arrays

Solution

1. Step 1: Understand greedy approach limitation

   Greedy picks the largest numbers first, which may overshoot or miss the exact target sum.

2. Step 2: How backtracking solves the problem

   Backtracking tries all subsets by including or excluding each number, ensuring it finds an exact sum if it exists.

3. Final Answer:

   Greedy fails because picking largest numbers first may miss exact sum; backtracking tries all subsets to find exact match -> Option A

4. Quick Check:

   Subset sum needs all subsets tried = Greedy fails because picking largest numbers first may miss exact sum; backtracking tries all subsets to find exact match [OK]

Hint: Subset sum needs all subsets; greedy picks only some [OK]

Common Mistakes:

• Thinking greedy always finds exact sum
• Confusing sorting with solving approach
• Believing backtracking is iterative only