Why Backtracking and What Greedy Cannot Solve in DSA Typescript - Challenge Your Understanding

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Backtracking Subset Sum

What is the output of the following TypeScript code that uses backtracking to find subsets summing to 5?

DSA Typescript

function backtrack(nums: number[], target: number, start: number, path: number[], res: number[][]): void {
  if (target === 0) {
    res.push([...path]);
    return;
  }
  for (let i = start; i < nums.length; i++) {
    if (nums[i] > target) continue;
    path.push(nums[i]);
    backtrack(nums, target - nums[i], i + 1, path, res);
    path.pop();
  }
}

const nums = [1, 2, 3, 4];
const res: number[][] = [];
backtrack(nums, 5, 0, [], res);
console.log(res);

A [[1,2,3],[1,4],[2,3]]

B [[1,4],[2,3]]

C [[1,4],[2,3],[5]]

D [[1,2,3,4]]

💡 Hint

Think about which combinations of numbers add exactly to 5 without repetition.

The backtracking function finds all unique subsets of nums that sum to 5. The valid subsets are [1,4] and [2,3]. The subset [1,2,3] sums to 6, so it's invalid. The number 5 is not in the list, so [5] is invalid. The full list sums to 10, so [1,2,3,4] is invalid.

🧠 Conceptual

intermediate

Why Greedy Fails for Subset Sum

Why does a greedy algorithm fail to solve the subset sum problem correctly in all cases?

A Because greedy algorithms make local optimal choices that may miss the correct subset sum.

B Because greedy algorithms always find the global optimum by checking all subsets.

C Because greedy algorithms use backtracking to explore all possibilities.

D Because greedy algorithms sort the input and always find the correct subset.

💡 Hint

Think about how greedy algorithms decide step-by-step without revisiting choices.

Greedy algorithms pick the best immediate option without considering future consequences. This can cause them to miss subsets that sum to the target because they don't explore all combinations.

❓ Predict Output

advanced

Backtracking vs Greedy Output Comparison

What is the output of the following TypeScript code comparing greedy and backtracking approaches for target sum 7?

DSA Typescript

function greedy(nums: number[], target: number): number[] {
  const res: number[] = [];
  let sum = 0;
  for (const num of nums) {
    if (sum + num <= target) {
      res.push(num);
      sum += num;
    }
  }
  return res;
}

function backtrack(nums: number[], target: number, start: number, path: number[], res: number[][]): void {
  if (target === 0) {
    res.push([...path]);
    return;
  }
  for (let i = start; i < nums.length; i++) {
    if (nums[i] > target) continue;
    path.push(nums[i]);
    backtrack(nums, target - nums[i], i + 1, path, res);
    path.pop();
  }
}

const nums = [2, 3, 5, 6];
const greedyResult = greedy(nums, 7);
const backtrackResult: number[][] = [];
backtrack(nums, 7, 0, [], backtrackResult);
console.log({greedyResult, backtrackResult});

A {"greedyResult":[2,3],"backtrackResult":[[2,5],[3,4]]}

B {"greedyResult":[2,3,5],"backtrackResult":[[2,5],[3,5]]}

C {"greedyResult":[2,3],"backtrackResult":[[2,5]]}

D }]]4,3[,]5,2[[:"tluseRkcartkcab",]3,2[:"tluseRydeerg"{

💡 Hint

Check which subsets sum exactly to 7 and what greedy picks.

The greedy picks 2, then 3, then tries 5 but sum would be 10 > 7, so stops at [2,3]. Backtracking finds [2,5] as the only subset summing to 7. The option with [3,4] is invalid because 4 is not in the list.

🔧 Debug

advanced

Identify the Error in Greedy Algorithm for Coin Change

What error does the following greedy coin change code produce when trying to make change for 6 with coins [1,3,4]?

DSA Typescript

function greedyCoinChange(coins: number[], amount: number): number[] {
  coins.sort((a, b) => b - a);
  const result: number[] = [];
  for (const coin of coins) {
    while (amount >= coin) {
      result.push(coin);
      amount -= coin;
    }
  }
  if (amount !== 0) throw new Error('Cannot make exact change');
  return result;
}

console.log(greedyCoinChange([1,3,4], 6));

A Returns [4,1,1] which sums to 6

B Returns [4,3] which sums to 7

C Returns [3,3] which sums to 6

D Throws 'Cannot make exact change' error

💡 Hint

Check how the greedy picks coins starting from largest.

The greedy picks 4 first, then two 1s to make 6. It returns [4,1,1]. It does not throw error because amount reaches 0. It does not pick [3,3] because 4 is chosen first greedily.

🧠 Conceptual

expert

Why Backtracking is Needed for N-Queens Problem

Why is backtracking necessary to solve the N-Queens problem instead of a greedy approach?

A Because greedy algorithms can place queens anywhere without checking conflicts.

B Because backtracking always places queens in the first row only.

C Because greedy algorithms use recursion to explore all board positions.

D Because greedy algorithms cannot guarantee placing all queens without conflicts by local choices alone.

💡 Hint

Think about how placing one queen affects future placements.

The N-Queens problem requires checking all possible placements to avoid conflicts. Greedy algorithms make local choices that may block future queen placements, so backtracking is needed to explore and undo choices.

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