Generate All Combinations Sum K in DSA Java - Deep Dive

Overview - Generate All Combinations Sum K

What is it?

Generating all combinations that sum to a target number K means finding every possible group of numbers from a given list that add up exactly to K. Each combination uses numbers without repetition and order does not matter. This helps explore all ways to reach a total using parts of a set.

Why it matters

Without this concept, solving problems like splitting bills, selecting items within a budget, or finding subsets with a specific sum would be hard or inefficient. It helps computers explore all valid options quickly and systematically, which is important in planning, optimization, and decision-making.

Where it fits

Before this, learners should understand arrays, recursion, and basic backtracking. After mastering this, they can explore more complex subset problems, dynamic programming for sums, and optimization algorithms.

Mental Model

Core Idea

Find every unique group of numbers from a list that add up exactly to a target sum by exploring choices step-by-step and backtracking when needed.

Think of it like...

It's like trying different combinations of coins from your pocket to pay exactly a certain amount, trying one coin at a time, and if it doesn't work, putting it back and trying another.

Start with empty set
  ↓
Pick a number -> Add to current sum
  ↓
If sum == K -> Save combination
If sum > K -> Backtrack
Else -> Pick next number
  ↓
Repeat until all numbers tried

Build-Up - 6 Steps

1

Foundation

Understanding the Problem Setup

Concept: Introduce the problem of finding combinations that sum to K from a list of numbers.

Given an array of positive integers and a target sum K, we want to find all unique combinations where the numbers add up to K. Each number can be used once per combination. Order does not matter, so [2,3] and [3,2] are the same combination.

Result

Clear problem statement and understanding of inputs and outputs.

Understanding the problem clearly is the first step to solving it correctly and efficiently.

2

Foundation

Basics of Backtracking Technique

Concept: Learn how to explore all possible combinations by choosing or skipping elements recursively.

Backtracking means trying a choice, exploring deeper, and if it doesn't lead to a solution, undoing the choice and trying another. For combinations sum, we pick a number, add it to current sum, and recurse. If sum exceeds K, we stop exploring that path.

Result

Ability to write a simple recursive function that tries all subsets.

Backtracking lets us systematically explore all options without missing or repeating combinations.

3

Intermediate

Avoiding Duplicate Combinations

🤔Before reading on: do you think sorting the array helps avoid duplicates or is it unnecessary? Commit to your answer.

Concept: Use sorting and skipping duplicates to ensure unique combinations only.

Sort the array first. When iterating, if the current number is the same as the previous one at the same recursion level, skip it. This prevents generating the same combination multiple times.

Result

Unique combinations without repeats even if input has duplicates.

Sorting combined with skipping duplicates is a simple and effective way to avoid repeated results.

4

Intermediate

Implementing Recursive Combination Builder

🤔Before reading on: do you think the recursion should stop when sum equals K or also when sum exceeds K? Commit to your answer.

Concept: Build a recursive function that tracks current combination and sum, adding valid combinations to results.

Create a helper function with parameters: current index, current combination list, and current sum. If sum == K, add combination to results. If sum > K, return early. Otherwise, loop from current index to end, pick number, recurse with updated sum and combination, then backtrack.

Result

Complete recursive solution that finds all valid combinations.

Stopping early when sum exceeds K saves time and avoids unnecessary work.

5

Advanced

Java Code with Backtracking and Pruning

🤔Before reading on: do you think using a List for current combination is better than an array? Commit to your answer.

Concept: Implement the full Java code using ArrayList for combinations and pruning to optimize.

import java.util.*; public class CombinationSumK { public List> combinationSum(int[] nums, int target) { List> results = new ArrayList<>(); Arrays.sort(nums); backtrack(nums, target, 0, new ArrayList<>(), results); return results; } private void backtrack(int[] nums, int target, int start, List current, List> results) { if (target == 0) { results.add(new ArrayList<>(current)); return; } for (int i = start; i < nums.length; i++) { if (i > start && nums[i] == nums[i - 1]) continue; // skip duplicates if (nums[i] > target) break; // prune current.add(nums[i]); backtrack(nums, target - nums[i], i + 1, current, results); current.remove(current.size() - 1); // backtrack } } } // Example usage: // int[] nums = {10,1,2,7,6,1,5}; // int target = 8; // CombinationSumK solver = new CombinationSumK(); // List> combos = solver.combinationSum(nums, target); // combos.forEach(System.out::println);

Result

[[1,1,6],[1,2,5],[1,7],[2,6]]

Using ArrayList allows dynamic combination building, and pruning reduces unnecessary recursion, improving efficiency.

6

Expert

Handling Large Inputs and Optimization

🤔Before reading on: do you think memoization can help in this problem or not? Commit to your answer.

Concept: Explore optimization techniques like pruning, sorting, and discuss why memoization is less effective here.

Memoization stores results of subproblems but here each path is unique due to index progression and combination state. Sorting and pruning are the main optimizations. For very large inputs, iterative solutions or dynamic programming variants can be considered but may not list all combinations. Parallel processing can speed up exploration.

Result

Understanding limits of backtracking and when to switch strategies.

Knowing when memoization helps and when it doesn't prevents wasted effort and guides choosing the right approach.

Under the Hood

The algorithm uses recursion to explore each number starting from a given index. It adds the number to the current combination and reduces the target sum accordingly. If the sum reaches zero, it records the combination. If the sum exceeds the target, it backtracks immediately. Sorting the array allows skipping duplicates and pruning branches early, reducing the search space.

Why designed this way?

Backtracking was chosen because it naturally fits exploring all possible subsets without repetition. Sorting and skipping duplicates prevent redundant work and repeated results. Alternatives like dynamic programming can find counts or existence but not list all unique combinations easily.

Start
  │
  ├─ Pick nums[0]
  │    ├─ sum + nums[0] == K -> Save
  │    ├─ sum + nums[0] < K -> Recurse next index
  │    └─ sum + nums[0] > K -> Backtrack
  ├─ Skip nums[0]
  │    └─ Recurse next index
  └─ Repeat for all indices

Duplicates skipped if nums[i] == nums[i-1] at same level

Myth Busters - 4 Common Misconceptions

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

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

Reality: Order does not matter; [2,3] and [3,2] represent the same combination and should be counted once.

Why it matters: Counting order as different leads to duplicate results and incorrect answers.

Quick: Do you think you can reuse the same number multiple times in one combination? Commit to yes or no.

Common Belief: You can reuse the same number multiple times in one combination.

Reality: Each number can be used only once per combination in this problem variant.

Why it matters: Reusing numbers changes the problem to a different variant and leads to wrong combinations.

Quick: Do you think sorting the input array is optional? Commit to yes or no.

Common Belief: Sorting the input array is optional and does not affect the result.

Reality: Sorting is essential to efficiently skip duplicates and prune search branches.

Why it matters: Without sorting, duplicates appear multiple times and performance degrades.

Quick: Do you think memoization can speed up generating all combinations? Commit to yes or no.

Common Belief: Memoization can speed up generating all combinations by caching results.

Reality: Memoization is less effective here because each path is unique due to index and combination state.

Why it matters: Trying to memoize wastes memory and complicates code without real speed gain.

Expert Zone

1

Skipping duplicates only works correctly if the array is sorted and skipping happens at the same recursion depth.

2

Backtracking uses the call stack implicitly; understanding stack depth helps avoid stack overflow on large inputs.

3

Pruning by breaking early when current number exceeds remaining target drastically reduces runtime in practice.

When NOT to use

Avoid this approach when the input size is huge and only the count of combinations is needed; use dynamic programming instead. Also, if numbers can be reused unlimited times, use a different variant of combination sum.

Production Patterns

Used in recommendation engines to find item bundles within budget, in finance to find investment portfolios matching target returns, and in testing to generate test cases covering all sum scenarios.

Connections

Subset Sum Problem

Builds-on

Understanding combination sum deepens grasp of subset sum, which asks if any subset sums to target, focusing on existence rather than listing all.

Backtracking Algorithm

Same pattern

Combination sum is a classic example of backtracking, showing how to explore solution spaces by making choices and undoing them.

Knapsack Problem (Optimization)

Related problem

Both involve selecting items under constraints; combination sum lists all valid selections, knapsack finds the best selection by value.

Common Pitfalls

#1 Including duplicate combinations due to not sorting or skipping duplicates.

Wrong approach:

for (int i = start; i < nums.length; i++) {
current.add(nums[i]);
backtrack(nums, target - nums[i], i + 1, current, results);
current.remove(current.size() - 1);
}

Correct approach:

for (int i = start; i < nums.length; i++) {
if (i > start && nums[i] == nums[i - 1]) continue;
current.add(nums[i]);
backtrack(nums, target - nums[i], i + 1, current, results);
current.remove(current.size() - 1);
}

Root cause:

Not checking for duplicates at the same recursion level causes repeated combinations.

#2 Continuing recursion even when current sum exceeds target, wasting time.

Wrong approach:

if (target == 0) {
results.add(new ArrayList<>(current));
return;
}
for (int i = start; i < nums.length; i++) {
current.add(nums[i]);
backtrack(nums, target - nums[i], i + 1, current, results);
current.remove(current.size() - 1);
}

Correct approach:

if (target == 0) {
results.add(new ArrayList<>(current));
return;
}
for (int i = start; i < nums.length; i++) {
if (nums[i] > target) break;
current.add(nums[i]);
backtrack(nums, target - nums[i], i + 1, current, results);
current.remove(current.size() - 1);
}

Root cause:

Not pruning branches when sum exceeds target leads to unnecessary recursion.

#3 Using a fixed-size array for current combination without tracking size, causing incorrect results.

Wrong approach:

int[] current = new int[nums.length];
// adding elements without managing current size
current[index] = nums[i];
backtrack(..., index + 1, ...);

Correct approach:

Use a dynamic list like ArrayList to add and remove elements cleanly:
List current = new ArrayList<>();
current.add(nums[i]);
backtrack(...);
current.remove(current.size() - 1);

Root cause:

Fixed arrays lack flexibility to track current combination length during recursion.

Key Takeaways

Generating all combinations that sum to K requires exploring choices recursively and backtracking when sums exceed the target.

Sorting the input and skipping duplicates at the same recursion level ensures unique combinations without repeats.

Pruning recursion early when the current number exceeds the remaining target saves time and improves efficiency.

Backtracking is a powerful technique to systematically explore all possible subsets or combinations.

Understanding problem constraints like single-use numbers and order irrelevance is crucial to correct implementation.

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 sum of all numbers from 1 to K

D. The product of all numbers that divide K

Solution

1. Step 1: Understand the problem goal

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

2. Step 2: Clarify what combinations mean

   Combinations mean selecting numbers without caring about order, but each combination must sum to K.

3. Final Answer:

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

4. Quick Check:

   Sum of combinations = K [OK]

Hint: Focus on sums equal to K, not order or multiplication [OK]

Common Mistakes:

• Confusing combinations with permutations
• Looking for product instead of sum
• Ignoring uniqueness of combinations

2. Which of the following Java method signatures is correct for a function that generates all combinations summing to K using backtracking?

easy

A. public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start)

B. public int findCombinations(int[] nums, int target)

C. void findCombinations(int target, int[] nums, List result)

D. public List<Integer> findCombinations(int target, int[] nums)

Solution

1. Step 1: Identify parameters needed for backtracking

   We need the array, target sum, current combination list, result list, and start index to avoid duplicates.

2. Step 2: Match method signature with these parameters

   public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) matches all needed parameters and returns void as it fills the result list.

3. Final Answer:

   public void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) -> Option A

4. Quick Check:

   Correct parameters for backtracking method [OK]

Hint: Backtracking needs current, result lists, and start index [OK]

Common Mistakes:

• Missing start index parameter
• Wrong return type for backtracking method
• Incorrect parameter order

3. What will be the output of the following Java code snippet?

List result = new ArrayList<>();
int[] nums = {2, 3, 6, 7};
int target = 7;
findCombinations(nums, target, new ArrayList<>(), result, 0);
System.out.println(result);

Assuming findCombinations correctly finds all unique combinations summing to target.

medium

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

B. [[7]]

C. [[2, 2, 3], [7]]

D. [[2, 2, 2, 3], [7]]

Solution

1. Step 1: Identify combinations that sum to 7

   From nums, 7 alone sums to 7, and 2+2+3 also sums to 7.

2. Step 2: Confirm no other combinations sum to 7

   3+6=9, 2+3+6=11, 2+2+2+3=9 (too large), so only [2,2,3] and [7] are valid.

3. Final Answer:

   [[2, 2, 3], [7]] -> Option C

4. Quick Check:

   Valid combinations sum to 7 [OK]

Hint: Check sums carefully; only valid combos appear [OK]

Common Mistakes:

• Including invalid sums
• Missing repeated elements allowed
• Confusing order of elements

4. Identify the error in this Java backtracking code snippet for generating combinations summing to K:

void findCombinations(int[] nums, int target, List<Integer> current, List result, int start) {
    if (target == 0) {
        result.add(new ArrayList<>(current));
        return;
    }
    for (int i = start; i < nums.length; i++) {
        if (nums[i] > target) continue;
        current.add(nums[i]);
        findCombinations(nums, target - nums[i], current, result, i);
        current.remove(current.size());
    }
}

medium

A. The recursive call should use i + 1 instead of i

B. The base case should check if target < 0 instead of target == 0

C. The loop should start from 0 instead of start

D. Removing element with current.remove(current.size()) causes IndexOutOfBoundsException

Solution

1. Step 1: Check the remove method usage

   List indices start at 0, so last element index is current.size() - 1, not current.size().

2. Step 2: Identify correct remove call

   Using current.remove(current.size()) causes IndexOutOfBoundsException because index is out of range.

3. Final Answer:

   Removing element with current.remove(current.size()) causes IndexOutOfBoundsException -> Option D

4. Quick Check:

   List remove index must be size-1 [OK]

Hint: Remove last element with index size-1, not size [OK]

Common Mistakes:

• Removing element at wrong index
• Ignoring base case correctness
• Wrong loop start index

5. Given the array nums = {1, 2, 3} and target sum 4, which of the following is the correct set of all unique combinations where numbers can be repeated and sum to 4?

hard

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

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

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

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

Solution

1. Step 1: List all combinations summing to 4 with repeats allowed

   Possible combos: [1,1,1,1], [1,1,2], [1,3], [2,2]. Note [3,1] is same as [1,3] and duplicates are not allowed.

2. Step 2: Check options for duplicates and invalid elements

   [[1,1,1,1], [1,1,2], [1,3], [2,2], [3,1]] includes [3,1] duplicate of [1,3]. [[1,1,1,1], [1,1,2], [1,3], [2,2], [4]] includes [4] which is not in nums. [[1,1,1,1], [1,1,2], [2,2], [3,1]] misses [1,3]. [[1,1,1,1], [1,1,2], [1,3], [2,2]] correctly lists unique combinations.

3. Final Answer:

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

4. Quick Check:

   Unique combos sum to 4 without duplicates [OK]

Hint: Avoid duplicates by sorting or fixed order in combos [OK]

Common Mistakes:

• Counting duplicates as separate combinations
• Including numbers not in the array
• Ignoring repeated use of numbers