KSum
- mosesg1123
- May 28
- 5 min read
The KSum problem generalizes 2Sum, 3Sum, and 4Sum into a flexible recursive solution that can find all unique combinations of k numbers in an array that sum to a target. It’s a fantastic problem for practicing recursion, backtracking, and thinking about how to build modular solutions that scale with complexity. Once you’ve nailed KSum, you can solve any of the others as special cases.
Problem Statement
Given an array of integers nums, an integer target, and an integer k, return all unique combinations of k numbers from nums that add up to target.
The result must not contain duplicate combinations.
Example:
Input: nums = [1, 0, -1, 0, -2, 2], target = 0, k = 4
Output: [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]1. Clarify Requirements Before Jumping Into Code
We're given a list of integers, a target value, and a number k. We need to find all unique k-length combinations that sum to the target.
Key clarifications:
Elements in the result must not repeat the same set (order doesn’t matter).
Input can have duplicate values, but we must avoid duplicate combinations in the output.
Result order doesn’t matter, but each combination should be sorted or deduplicated internally.
2. Identify the Category of the Question
This is a generalized sum combination problem.
It fits in the recursive/backtracking + two pointers family. It builds upon:
2Sum (two pointers)
3Sum (loop + 2Sum)
4Sum (nested loop + 2Sum)
Common patterns:
Recursion
Sorting to eliminate duplicates
Two pointers for base case (k = 2)
Pruning (early exits when the smallest/largest values can’t help)
3. Consider a Brute‑Force Approach to Understand the Problem
Brute-force would be generating all possible k-length subsets and filtering ones with the desired sum.
Time complexity:
Generating subsets: O(n^k)
Checking for duplicates: adds more overhead
We can do better, but it helps reinforce that we need:
Pruning
Efficient subset generation
A smarter base case (like 2Sum with two pointers)
4. Brainstorm More Solutions
This is building on a set of problems that we've seen before. So, like we did with the 4Sum problem and the 3Sum Closest problem, let's start by reviewing the solution for the 3Sum problem:
Step 1: Sort the array.
Step 2: Loop through each element i as the first number of the triplet.
Step 3: For each element i, use two pointers:
left = i + 1
right = n - 1
While left < right, compute the current sum of the three numbers.
If it’s equal to the target, store it.
If the sum is less than the target, move the left pointer to increase the sum.
If the sum is greater than the target, move the right pointer to decrease the sum.
Let's also remember that for the 4Sum solution, we built on the solution for the 3Sum problem by fixing two indices (i and j) instead of just one index (i). Then we applied the same two-pointer pattern we discovered way back in the 2Sum problem, scanning the remaining subarray.
What's the common pattern we're seeing throughout all these problems? In all solutions, we are fixing as many elements as necessary (one for the 3Sum problem, two for the 4Sum problem) so that we can simplify the problem down to where we can use the two-pointer pattern to scan the remaining subarray.
If we want to generalize that to k sums, we probably want to continue using this pattern in some way. But, we can't do it with loops like we did in the previous solutions, because we have no idea how many loops to use.
Can we design an algorithm that will break down the problem into smaller sub-problems until I get down to a version of the problem that I can implement with a simple loop like the 2Sum solution? There's one technique that fits that description - recursion! It would give us a runtime of O(n^(k-1)) , which is right in line with the runtimes that we are able to achieve with the 3Sum and 4Sum problems.
So to summarize the logic here:
Sort the array to simplify deduplication
Write a recursive kSum(nums, start, k, target) function
Base case: when k == 2, use two pointers to find pairs that sum to target
Recursive case: fix one element and recurse with k-1
5. Discuss Trade‑Offs Between Your Solutions
Approach | Time Complexity | Space Complexity | Pros | Cons |
Brute Force | O(n^k) | O(k) | Simple but unscalable | Fails for n > 20 or large k |
Recursive KSum | O(n^(k-1)) | O(k) | Elegant, reusable, scalable | Needs careful pruning and dedup |
6. Write Pseudocode to Structure Your Thoughts
function kSum(nums, start, k, target):
result = []
if k == 2:
use two pointers to find pairs from start to end
skip duplicates
add valid pairs to result
else:
for i from start to len(nums) - k:
if i > start and nums[i] == nums[i-1]: continue
if nums[i] * k > target: break
if nums[-1] * k < target: break
recursive = kSum(nums, i+1, k-1, target - nums[i])
for each comb in recursive:
add [nums[i]] + comb to result
return result
7. Consider Edge Cases
Length of array < k → return empty list
Large positive or negative numbers may cause overflow if not careful
Large arrays may cause stack overflow
Target impossible to reach given min/max possible sums
8. Write Full Code Syntax
public class KSumSolver {
public List<List<Integer>> kSum(int[] nums, int k, int target) {
Arrays.sort(nums);
return kSumHelper(nums, 0, k, target);
}
private List<List<Integer>> kSumHelper(int[] nums, int start, int k, int target) {
List<List<Integer>> res = new ArrayList<>();
int n = nums.length;
if (k == 2) {
int left = start, right = n - 1;
while (left < right) {
int sum = nums[left] + nums[right];
if (sum == target) {
res.add(Arrays.asList(nums[left], nums[right]));
left++; right--;
while (left < right && nums[left] == nums[left - 1]) left++;
while (left < right && nums[right] == nums[right + 1]) right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
} else {
for (int i = start; i < n - k + 1; i++) {
if (i > start && nums[i] == nums[i - 1]) continue;
if ((long) nums[i] + (long) nums[n - 1] * (k - 1) < target) continue;
if ((long) nums[i] * k > target) break;
List<List<Integer>> temp = kSumHelper(nums, i + 1, k - 1, target - nums[i]);
for (List<Integer> t : temp) {
List<Integer> curr = new ArrayList<>();
curr.add(nums[i]);
curr.addAll(t);
res.add(curr);
}
}
}
return res;
}
}
9. Test Your Code
public static void main(String[] args) {
KSumSolver solver = new KSumSolver();
System.out.println(solver.kSum(new int[]{1, 0, -1, 0, -2, 2}, 4, 0));
// Expected: [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
System.out.println(solver.kSum(new int[]{2, 2, 2, 2, 2}, 4, 8));
// Expected: [[2, 2, 2, 2]]
System.out.println(solver.kSum(new int[]{-1, 0, 1, 2, -1, -4}, 3, 0));
// Expected: [[-1, -1, 2], [-1, 0, 1]]
System.out.println(solver.kSum(new int[]{1000000000, 1000000000, 1000000000, 1000000000}, 4, 4000000000L));
// Expected: [[1000000000, 1000000000, 1000000000, 1000000000]]
System.out.println(solver.kSum(new int[]{}, 4, 0));
// Expected: []
}
10. Key Lessons to Remember for Future Questions
KSum builds on the logic of 2Sum and 3Sum using recursion.
Sorting is essential for both deduplication and pruning.
When designing recursive algorithms, base case generalization (like 2Sum) helps build elegant and scalable solutions.
Watch out for overflows with large integers—use long types when summing.
You can use this template to build a reusable utility function for real-world apps requiring complex combination logic.
Good Luck and Happy Coding!
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