kSum Problem
Problem
Given an array S of n integers, are there elements a1
, a2
, a3
…ak in
S such that sum(a1, a2, a3, ... ak) = target
? Find all unique lists in the array which gives the sum of target.
Examples
Refer problems like 4Sum, 3Sum - Classic Problem, 3Sum0 - Find three elements in an array that sum to a zero for examples.
Solution
Method 1 - Backtracking
public List<List<Integer>> kSum(int[] nums, int target, int k) {
len = nums.length;
Arrays.sort(nums);
return kSum(nums, target, k, 0);
}
private List<List<Integer>> kSum(int[] nums, int target, int k, int index) {
List<List<Integer>> res = new ArrayList<>();
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);
//skip duplication
while(i<j && nums[i]==nums[i+1]) i++;
while(i<j && nums[j-1]==nums[j]) j--;
i++;
j--;
//move left bound
} else if (target - nums[i] > nums[j]) {
i++;
//move right bound
} else {
j--;
}
}
} else{
for (int i = index; i < len - k + 1; i++) {
//use current number to reduce ksum into k-1sum
List<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);
}
while (i < len-1 && nums[i] == nums[i+1]) {
//skip duplicated numbers
i++;
}
}
}
return res;
}
Complexity
- ⏰ Time complexity:
O(n^(k-1))
- 🧺 Space complexity:
O(1)