Generate All Combinations Sum K in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the recursive function for combinations.

DSA Javascript

function combineSumK(nums, k) {
  const result = [];
  function backtrack(start, path, sum) {
    if (sum === k) {
      result.push([...path]);
      return;
    }
    for (let i = [1]; i < nums.length; i++) {
      if (sum + nums[i] > k) continue;
      path.push(nums[i]);
      backtrack(i + 1, path, sum + nums[i]);
      path.pop();
    }
  }
  backtrack(0, [], 0);
  return result;
}

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

Anums[0]

B1

Cstart

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 1 skips the first element.

Using sum or nums[0] as start index causes errors.

The loop should start from 'start' to consider all elements from the current position.

2fill in blank

medium

Complete the code to check if the current sum equals the target k.

DSA Javascript

function backtrack(start, path, sum) {
  if ([1] === k) {
    result.push([...path]);
    return;
  }
  // rest of code
}

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

Apath.length

Bsum

Cstart

Dnums.length

Attempts:

3 left

💡 Hint

Common Mistakes

Checking path length instead of sum.

Comparing start or nums.length to k.

We compare the current sum with k to find valid combinations.

3fill in blank

hard

Fix the error in the pruning condition to skip sums greater than k.

DSA Javascript

for (let i = start; i < nums.length; i++) {
  if ([1] + nums[i] > k) continue;
  path.push(nums[i]);
  backtrack(i + 1, path, sum + nums[i]);
  path.pop();
}

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

Asum

Bstart

Ci

Dpath.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using start or i instead of sum causes wrong pruning.

Using path length instead of sum.

We add nums[i] to the current sum to check if it exceeds k.

4fill in blank

hard

Fill in the blanks to correctly call the backtrack function and remove the last element.

DSA Javascript

path.push(nums[i]);
backtrack([1], path, [2]);
path.[3]();

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

Ai + 1

Bsum + nums[i]

Cpop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using push instead of pop to remove element.

Not updating sum correctly in recursive call.

We move to the next index i+1, update sum by adding nums[i], and remove last element with pop after recursion.

5fill in blank

hard

Fill all three blanks to complete the function that returns all combinations summing to k.

DSA Javascript

function combineSumK(nums, k) {
  const result = [];
  function backtrack([1], path, [2]) {
    if ([2] === k) {
      result.push([...path]);
      return;
    }
    for (let i = [1]; i < nums.length; i++) {
      if ([2] + nums[i] > k) continue;
      path.push(nums[i]);
      backtrack(i + 1, path, [2] + nums[i]);
      path.pop();
    }
  }
  backtrack(0, [], 0);
  return result;
}

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

Astart

Bsum

C0

Dpath

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names for parameters.

Starting loop from 0 instead of 'start' inside backtrack.

The backtrack function uses 'start' as the first parameter, 'sum' as the third parameter, and the loop starts from 'start'.

Practice

1. What does the problem "Generate All Combinations Sum K" ask you to find?

easy

A. All unique combinations of numbers that add up to K

B. The largest number less than K

C. The count of numbers greater than K

D. The product of all numbers up to K

Solution

1. Step 1: Understand the problem goal

   The problem asks to find all unique sets of numbers that sum exactly to K.

2. Step 2: Clarify what combinations mean

   Combinations mean groups of numbers without regard to order, all adding to K.

3. Final Answer:

   All unique combinations of numbers that add up to K -> Option A

4. Quick Check:

   Combinations sum to K = All unique combinations of numbers that add up to K [OK]

Hint: Focus on sums equal to K, not counts or products [OK]

Common Mistakes:

• Confusing combinations with permutations
• Looking for counts instead of actual combinations
• Trying to find products instead of sums

2. Which of the following is the correct way to start a recursive function in JavaScript to generate combinations summing to K?

easy

A. function findCombos() { return; }

B. function findCombos(nums, target) { return nums + target; }

C. function findCombos(nums, target, start, path, result) { ... }

D. function findCombos(nums) { console.log(target); }

Solution

1. Step 1: Identify correct recursive function signature

   The function needs parameters for numbers, target sum, start index, current path, and result storage.

2. Step 2: Check options for these parameters

   Only function findCombos(nums, target, start, path, result) { ... } includes all necessary parameters for recursion and backtracking.

3. Final Answer:

   function findCombos(nums, target, start, path, result) { ... } -> Option C

4. Quick Check:

   Recursive function with all params = function findCombos(nums, target, start, path, result) { ... } [OK]

Hint: Include start index and path for backtracking recursion [OK]

Common Mistakes:

• Missing parameters needed for recursion
• Returning wrong values instead of building combinations
• Not including start index to avoid duplicates

3. What is the output of the following code snippet?

function combinationSum(nums, target) {
  const result = [];
  function backtrack(start, path, sum) {
    if (sum === target) {
      result.push([...path]);
      return;
    }
    if (sum > target) return;
    for (let i = start; i < nums.length; i++) {
      path.push(nums[i]);
      backtrack(i + 1, path, sum + nums[i]);
      path.pop();
    }
  }
  backtrack(0, [], 0);
  return result;
}

console.log(combinationSum([2,3,6,7], 7));

medium

A. [[7]]

B. [[2,3,6,7]]

C. [[2,3,7]]

D. [[2,2,3]]

Solution

1. Step 1: Understand the code logic

   The function finds combinations of numbers from the array that sum to 7 without reuse (i + 1).

2. Step 2: Trace the combinations

   Only 7 alone sums to 7; others like 2+3=5, 2+3+6=11 exceed or don't match target.

3. Final Answer:

   [[7]] -> Option A

4. Quick Check:

   Only [7] sums to 7 without repeats = [[7]] [OK]

Hint: Check sum and index increment to avoid reuse [OK]

Common Mistakes:

• Assuming numbers can be reused multiple times
• Including combinations that exceed target
• Not copying path array before pushing

4. Identify the bug in this code snippet for generating combinations summing to K:

function findCombos(nums, target) {
  const result = [];
  function backtrack(start, path, sum) {
    if (sum === target) {
      result.push(path);
      return;
    }
    if (sum > target) return;
    for (let i = start; i < nums.length; i++) {
      path.push(nums[i]);
      backtrack(i + 1, path, sum + nums[i]);
      path.pop();
    }
  }
  backtrack(0, [], 0);
  return result;
}

medium

A. The base case sum === target is missing

B. The path array is pushed directly without copying, causing reference issues

C. The loop should start from 0 instead of start

D. The function does not return any result

Solution

1. Step 1: Check how results are stored

   The code pushes 'path' directly, which is a reference and will change later.

2. Step 2: Identify correct way to store combinations

   Must push a copy of path (e.g., [...path]) to save current state.

3. Final Answer:

   The path array is pushed directly without copying, causing reference issues -> Option B

4. Quick Check:

   Push copy of path, not reference = The path array is pushed directly without copying, causing reference issues [OK]

Hint: Always push a copy of path array to result [OK]

Common Mistakes:

• Pushing path reference instead of copy
• Incorrect loop start index causing duplicates
• Missing base case for sum equal to target

5. Given the array [1, 2, 3, 4] and target sum 5, which of the following is the correct set of all unique combinations that sum to 5, assuming numbers can be used only once?

hard

A. [[1,2,2]]

B. [[1,1,3],[2,3]]

C. [[5]]

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

Solution

1. Step 1: List all combinations without reuse that sum to 5

   Possible pairs: 1+4=5, 2+3=5; triples or singles don't sum to 5 without reuse.

2. Step 2: Verify options against these combinations

   [[1,4],[2,3]] matches these pairs exactly; others include duplicates or invalid sums.

3. Final Answer:

   [[1,4],[2,3]] -> Option D

4. Quick Check:

   Unique pairs summing to 5 = [[1,4],[2,3]] [OK]

Hint: Sum pairs without reuse to find valid combos [OK]

Common Mistakes:

• Including repeated numbers when only single use allowed
• Assuming single number 5 exists in array
• Counting invalid sums or duplicates