Generate All Combinations Sum K in DSA Typescript - Deep Dive

Overview - Generate All Combinations Sum K

What is it?

Generate All Combinations Sum K means finding all unique groups of numbers from a list that add up exactly to a target number K. Each number can be used once or multiple times depending on the problem rules. The goal is to list every possible combination that meets this sum. This helps solve problems where you want to explore all ways to reach a total using given parts.

Why it matters

Without this concept, solving problems that require exploring all possible sums would be slow and error-prone. It helps in budgeting, resource allocation, and puzzle solving where you need to find all ways to combine items to reach a goal. It also builds a foundation for understanding backtracking and recursion, which are key in many complex algorithms.

Where it fits

Before learning this, you should understand arrays, basic loops, and recursion. After mastering this, you can explore more complex backtracking problems, dynamic programming, and optimization techniques.

Mental Model

Core Idea

Find all groups of numbers that add up to K by trying each number and exploring further only if the sum stays valid.

Think of it like...

It's like trying different combinations of ingredients to bake a cake that weighs exactly K grams, testing each ingredient amount and backtracking if the total weight goes over.

Start
  ↓
Choose a number -> Add to current sum
  ↓               ↓
If sum == K -> Save combination
If sum > K -> Backtrack
Else -> Try next number
  ↓
Repeat until all numbers tried

Build-Up - 7 Steps

1

Foundation

Understanding the Problem Setup

Concept: Learn what it means to find combinations that sum to K and the problem constraints.

We have a list of numbers and a target sum K. We want to find all unique sets of numbers from the list that add up exactly to K. Each number can be used once or multiple times depending on the problem variant. The output is a list of these sets.

Result

Clear understanding of the problem goal and input/output format.

Understanding the problem fully prevents confusion later and guides how to approach the solution.

2

Foundation

Basic Recursion for Sum Exploration

Concept: Use recursion to explore sums by adding numbers step-by-step.

Start with an empty combination and sum 0. For each number, add it to the combination and sum, then recursively try adding more numbers. If sum equals K, save the combination. If sum exceeds K, stop exploring that path.

Result

A recursive method that tries all possible sums by adding numbers.

Recursion naturally fits problems where you explore many possible choices step-by-step.

3

Intermediate

Avoiding Duplicate Combinations

🤔Before reading on: do you think sorting the list helps avoid duplicates or not? Commit to your answer.

Concept: Sort the input and skip repeated numbers during recursion to avoid duplicate combinations.

By sorting the list, duplicates appear next to each other. When choosing numbers, if the current number is the same as the previous one and the previous was not chosen in this recursion level, skip it. This prevents generating the same combination multiple times.

Result

Unique combinations without repeats in the output.

Sorting combined with skipping duplicates is a simple but powerful way to ensure uniqueness.

4

Intermediate

Controlling Number Usage Frequency

🤔Before reading on: do you think numbers can be reused unlimited times or only once? Commit to your answer.

Concept: Decide if numbers can be reused multiple times or only once and adjust recursion accordingly.

If numbers can be reused, after choosing a number, recurse with the same index to allow reuse. If only once, recurse with the next index to avoid reuse. This changes how the recursion tree grows.

Result

Correct combinations respecting usage rules.

Understanding usage rules changes recursion structure and output significantly.

5

Intermediate

Building the Combination Result

Concept: Keep track of the current combination and add it to results when sum matches K.

Use a temporary list to store chosen numbers. When sum equals K, copy this list and add it to the final results. Backtrack by removing the last number before trying the next choice.

Result

A list of all valid combinations collected.

Backtracking with temporary storage allows exploring all paths without losing progress.

6

Advanced

Efficient Pruning to Improve Performance

🤔Before reading on: do you think checking sum early helps speed or wastes time? Commit to your answer.

Concept: Stop exploring paths as soon as the sum exceeds K to avoid unnecessary work.

During recursion, if the current sum plus the next number exceeds K, stop exploring that branch. This pruning reduces the number of recursive calls and speeds up the algorithm.

Result

Faster execution with fewer recursive calls.

Pruning prevents wasted effort and is key for scaling to larger inputs.

7

Expert

Handling Large Inputs and Memory Optimization

🤔Before reading on: do you think storing all combinations is always feasible? Commit to your answer.

Concept: Use generators or yield results one by one to save memory when output is large.

Instead of storing all combinations in memory, yield each valid combination as soon as found. This allows processing results on the fly and reduces memory usage. Also, careful copying of combinations avoids unnecessary memory overhead.

Result

Memory-efficient solution suitable for large datasets.

Knowing how to manage memory and output format is crucial for real-world applications.

Under the Hood

The algorithm uses recursion and backtracking to explore all possible combinations. At each step, it chooses a number, adds it to the current sum and combination, then recursively explores further. If the sum matches the target, it records the combination. If the sum exceeds the target, it backtracks immediately. Sorting and skipping duplicates ensure unique results. The recursion tree represents all possible choices and pruning cuts off invalid branches early.

Why designed this way?

Backtracking was chosen because it naturally fits problems requiring exploration of all combinations without repetition. Sorting and skipping duplicates prevent redundant work and repeated results. Pruning improves efficiency by avoiding impossible paths. Alternatives like brute force would be too slow, and dynamic programming doesn't easily list all combinations, so backtracking balances completeness and performance.

Recursion Tree:

Root (sum=0, combo=[])
├─ Choose num1 -> sum=num1, combo=[num1]
│  ├─ Choose num1 again (if reuse allowed)
│  ├─ Choose num2
│  └─ ...
├─ Choose num2 -> sum=num2, combo=[num2]
│  ├─ ...
└─ ...

Pruning cuts branches where sum > K.

Myth Busters - 4 Common Misconceptions

Quick: Do you think the order of numbers in combinations matters? Commit yes or no.

Common Belief: The order of numbers in combinations matters, so [2,3] and [3,2] are different.

Reality: Order does not matter in combinations; [2,3] and [3,2] represent the same combination and should appear only once.

Why it matters: Treating order as different leads to duplicate results and confusion about the number of unique solutions.

Quick: Do you think you must try all numbers every time, even if sum already exceeds K? Commit yes or no.

Common Belief: You must try all numbers regardless of current sum to find all combinations.

Reality: If the current sum exceeds K, continuing to add numbers cannot produce valid combinations, so you should stop exploring that path immediately.

Why it matters: Not pruning wastes time and slows down the algorithm drastically on larger inputs.

Quick: Do you think duplicates in input always produce duplicates in output? Commit yes or no.

Common Belief: Duplicates in the input list always cause duplicate combinations in the output.

Reality: With proper sorting and skipping duplicates during recursion, duplicate input numbers do not cause duplicate output combinations.

Why it matters: Failing to handle duplicates leads to bloated output and incorrect answers.

Quick: Do you think backtracking always uses a lot of memory? Commit yes or no.

Common Belief: Backtracking always requires storing all combinations in memory at once.

Reality: Backtracking can be implemented to yield combinations one by one, reducing memory usage significantly.

Why it matters: Assuming high memory use may discourage using backtracking when it is actually efficient with proper implementation.

Expert Zone

1

Choosing whether to allow reuse of numbers changes the recursion index management subtly but critically.

2

Sorting input not only helps skip duplicates but also enables early pruning by stopping when numbers exceed the remaining sum.

3

Copying the current combination before saving is necessary to avoid mutation bugs, but excessive copying can hurt performance if not managed carefully.

When NOT to use

This approach is not suitable when the input size is extremely large and the number of combinations explodes exponentially. In such cases, approximate or heuristic methods, or dynamic programming counting without listing combinations, are better alternatives.

Production Patterns

Used in recommendation systems to find all item bundles matching a budget, in finance to find all investment combinations for a target return, and in puzzle games to generate all valid moves or solutions.

Connections

Backtracking

Builds-on

Understanding this problem deepens knowledge of backtracking, a fundamental technique for exploring all possible solutions in constraint problems.

Subset Sum Problem

Same pattern

This problem is a direct extension of the subset sum problem, where the goal is to find any subset summing to K, but here we find all such subsets.

Combinatorial Chemistry

Analogous process

In combinatorial chemistry, scientists combine different molecules to find all compounds with desired properties, similar to generating all number combinations summing to a target.

Common Pitfalls

#1 Not sorting input and skipping duplicates causes repeated combinations.

Wrong approach:

function backtrack(start, sum, combo) {
for (let i = start; i < nums.length; i++) {
combo.push(nums[i]);
backtrack(i + 1, sum + nums[i], combo);
combo.pop();
}
}

Correct approach:

nums.sort((a,b) => a-b);
function backtrack(start, sum, combo) {
for (let i = start; i < nums.length; i++) {
if (i > start && nums[i] === nums[i-1]) continue;
combo.push(nums[i]);
backtrack(i + 1, sum + nums[i], combo);
combo.pop();
}
}

Root cause:

Ignoring duplicates in input leads to exploring identical branches multiple times.

#2 Continuing recursion even when sum exceeds K wastes time.

Wrong approach:

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

Correct approach:

function backtrack(start, sum, combo) {
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, sum + nums[i], combo);
combo.pop();
}
}

Root cause:

Not checking sum before recursion causes unnecessary exploration.

#3 Modifying the combination array without copying causes wrong results.

Wrong approach:

if (sum === target) results.push(combo);

Correct approach:

if (sum === target) results.push([...combo]);

Root cause:

Storing references to a mutable array leads to all saved results changing together.

Key Takeaways

Generating all combinations that sum to K is a classic backtracking problem requiring careful exploration and pruning.

Sorting input and skipping duplicates ensures unique combinations without repeats.

Pruning branches where the sum exceeds K greatly improves efficiency.

Managing combination storage carefully avoids bugs and memory issues.

Understanding usage rules for numbers (reuse or not) changes the recursion logic and output.

Practice

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

easy

A. The count of numbers equal to K

B. The largest number less than K

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

D. The smallest number greater than K

Solution

1. Step 1: Understand the problem goal

   The problem asks to find all unique sets of numbers whose sum equals K.

2. Step 2: Clarify uniqueness and sum condition

   Each combination must sum exactly to K without duplicates in the result.

3. Final Answer:

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

4. Quick Check:

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

Hint: Focus on combinations that sum exactly to K [OK]

Common Mistakes:

• Confusing sum with count
• Including duplicate combinations
• Looking for single numbers only

2. Which of the following TypeScript function signatures correctly represents a function to generate all combinations summing to K from an array of numbers?

easy

A. function generateCombinations(nums: number, target: number[]): number[]

B. function generateCombinations(nums: number[], target: number): number[][]

C. function generateCombinations(nums: string[], target: string): string[]

D. function generateCombinations(nums: number[], target: number): number[]

Solution

1. Step 1: Analyze parameter types

   The function should take an array of numbers and a target number.

2. Step 2: Analyze return type

   It should return an array of arrays of numbers representing combinations.

3. Final Answer:

   function generateCombinations(nums: number[], target: number): number[][] -> Option B

4. Quick Check:

   Input array and target number, output array of number arrays [OK]

Hint: Input array and target number, output array of arrays [OK]

Common Mistakes:

• Using wrong parameter types
• Returning flat array instead of array of arrays
• Mixing string and number types

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

function combineSum(nums: number[], target: number): number[][] {
  const result: number[][] = [];
  function backtrack(start: number, path: number[], sum: number) {
    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(combineSum([1, 2, 3], 3));

medium

A. [[1, 2], [3]]

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

C. [[3], [2, 1]]

D. [[1, 2, 3]]

Solution

1. Step 1: Trace backtracking calls

   Starting from 0, combinations are built adding numbers without reuse: [1,2] sums to 3, [3] sums to 3.

2. Step 2: Confirm valid combinations

   Other combinations like [2] or [1,3] do not sum to 3 or are invalid due to index increment.

3. Final Answer:

   [[1, 2], [3]] -> Option A

4. Quick Check:

   Combinations summing to 3 = [[1, 2], [3]] [OK]

Hint: Check sums and index increments carefully [OK]

Common Mistakes:

• Including combinations with reused elements
• Ignoring sum > target condition
• Confusing order of elements

4. Identify the bug in this TypeScript function that aims to generate all combinations summing to K:

function findCombos(nums: number[], target: number): number[][] {
  const res: number[][] = [];
  function dfs(i: number, path: number[], total: number) {
    if (total === target) {
      res.push(path);
      return;
    }
    if (total > target || i === nums.length) return;
    dfs(i + 1, path, total);
    path.push(nums[i]);
    dfs(i + 1, path, total + nums[i]);
    path.pop();
  }
  dfs(0, [], 0);
  return res;
}

medium

A. The function does not return any result

B. The base case does not check for total > target

C. The recursion does not increment index i

D. The path array is pushed by reference causing incorrect results

Solution

1. Step 1: Check how path is stored

   The path array is pushed directly without copying, so later changes affect stored results.

2. Step 2: Understand impact on results

   Because path is referenced, all stored combinations end up empty or incorrect after backtracking.

3. Final Answer:

   The path array is pushed by reference causing incorrect results -> Option D

4. Quick Check:

   Push copy of path, not reference [OK]

Hint: Always push a copy of path, not the original [OK]

Common Mistakes:

• Pushing path without copying
• Not handling base cases properly
• Incorrect index increment

5. Given an array with duplicates, how can you modify the combination sum algorithm to generate all unique combinations summing to K without duplicates in the result?

nums = [1, 2, 2, 3], target = 4

hard

A. Sort the array and skip duplicates during recursion

B. Use a Set to store all combinations as strings

C. Allow duplicates and filter results after generation

D. Ignore duplicates and generate all combinations

Solution

1. Step 1: Sort the input array

   Sorting groups duplicates together, making it easier to skip them during recursion.

2. Step 2: Skip duplicates in recursion

   When iterating, skip elements equal to the previous to avoid duplicate combinations.

3. Final Answer:

   Sort the array and skip duplicates during recursion -> Option A

4. Quick Check:

   Sorting + skipping duplicates = unique combinations [OK]

Hint: Sort and skip duplicates while recursing [OK]

Common Mistakes:

• Not sorting before recursion
• Filtering duplicates after generation (inefficient)
• Ignoring duplicates causing repeated results