# K Sums

All K Sum problem can be divided into two problems:

* 2 Sum Problem 
* Reduce K Sum problem to K – 1 Sum Problem

从4Sum和3Sum，我们可以看出对于KSum的通用套路：将KSum转化为K-1 Sum，最后用2Sum的Two Pointer求解。

注意要点：**先排序**，**去除/跳过重复元素**

### Generalized K Sum Approach:

```java
class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        return kSum(nums, 0, 4, target);
    }

    private List<List<Integer>> kSum (int[] nums, int start, int k, int target) {
        int len = nums.length;
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (k == 2) { 
            //two pointers from left and right
            int left = start, right = len - 1;
            while (left < right) {
                int sum = nums[left] + nums[right];
                if (sum == target) {
                    List<Integer> path = new ArrayList<Integer>();
                    path.add(nums[left]);
                    path.add(nums[right]);
                    res.add(path);
                    left++;
                    right--;

                    while(left < right && nums[left] == nums[left - 1]) left++;
                    while(left < right && nums[right] == nums[right + 1]) right--;

                } else if (sum < target) { //move left
                    left++;
                } else { //move right
                    right--;
                }
            }
        } else {
            for(int i = start; i < len - (k - 1); i++) {
                if(i > start && nums[i] == nums[i - 1]) continue;
                List<List<Integer>> temp = kSum(nums, i + 1, k - 1, target - nums[i]);
                for(List<Integer> t : temp) {
                    t.add(0, nums[i]);
                }                    
                res.addAll(temp);
            }
        }
        return res;
    }
}
```

### Another K Sum

```java
public class Solution {
    int len = 0;
    public List<List<Integer>> fourSum(int[] nums, int target) {
        len = nums.length;
        Arrays.sort(nums);
        return kSum(nums, target, 4, 0);
    }
    private ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int index) {
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        if (index >= len) {
            return res;
        }
        if (k == 2) {
            int i = index, j = len - 1;
            while (i < j) {
                //find a pair
                if (target - nums[i] == nums[j]) {
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums[i]);
                    temp.add(target-nums[i]);
                    res.add(temp);
                    i++;
                    j--;
                    //skip duplication
                    while (i < j && nums[i] == nums[i - 1]) i++;
                    while (i < j && nums[j] == nums[j + 1]) j--;
                } else if (target - nums[i] > nums[j]) {
                    //move left bound
                    i++;
                } else {
                    //move right bound
                    j--;
                }
            }
        } else {
            for (int i = index; i < len - k + 1; i++) {
                // Alternative: Skipping duplicates in the front
                // if (i > index && nums[i] == nums[i - 1]) {
                //    continue;
                // }
                //use current number to reduce k sum into k-1 sum
                ArrayList<List<Integer>> temp = kSum(nums, target - nums[i], k - 1, i + 1);
                if (temp != null) {
                    //add previous results
                    for (List<Integer> t : temp) {
                        t.add(0, nums[i]);
                    }
                    res.addAll(temp);
                }
                //skip duplicated numbers
                while (i < len - 1 && nums[i] == nums[i + 1]) {
                    i++;
                }
            }
        }
        return res;
    }
}
```

## Reference

Generalized KSums: <https://leetcode.com/problems/4sum/discuss/8609/My-solution-generalized-for-kSums-in-JAVA>


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://aaronice.gitbook.io/lintcode/problem-solving-summary/k-sums.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.