Why Generate All Combinations Sum K in DSA Javascript?

The Big Idea

What if you could instantly find every way to combine numbers to hit your target without missing a single one?

The Scenario

Imagine you want to find all ways to pick numbers from a list that add up to a target sum, like choosing coins to pay exactly $10. Doing this by hand means trying every possible group, which quickly becomes confusing and slow.

The Problem

Manually checking every combination is tiring and easy to mess up. As the list grows, the number of groups to check grows even faster, making it almost impossible to do without missing some or repeating others.

The Solution

Using a smart method called backtracking, we can automatically explore all possible groups step-by-step, skipping groups that won't work early. This saves time and ensures we find all correct combinations without repeats.

Before vs After

✗ Before

const nums = [1,2,3];
// Manually try all combos
// 1+2=3, 1+3=4, 2+3=5, etc.
// Too many checks, easy to miss

✓ After

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

What It Enables

This lets you quickly find every possible group of numbers that add up to your target, no matter how big the list is.

Real Life Example

Planning a party budget where you want to combine different snack prices to exactly match your budget without overspending or missing options.

Key Takeaways

Manual checking is slow and error-prone for combination sums.

Backtracking explores all options efficiently and cleanly.

It helps find all groups that add up to a target sum without repeats.

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