Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
Java Solution
A typical k-sum problem. Time is N to the power of (k-1).
public List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> result = new ArrayList<List<Integer>>(); if(nums==null|| nums.length<4) return result; Arrays.sort(nums); for(int i=0; i<nums.length-3; i++){ if(i!=0 && nums[i]==nums[i-1]) continue; for(int j=i+1; j<nums.length-2; j++){ if(j!=i+1 && nums[j]==nums[j-1]) continue; int k=j+1; int l=nums.length-1; while(k<l){ if(nums[i]+nums[j]+nums[k]+nums[l]<target){ k++; }else if(nums[i]+nums[j]+nums[k]+nums[l]>target){ l--; }else{ List<Integer> t = new ArrayList<Integer>(); t.add(nums[i]); t.add(nums[j]); t.add(nums[k]); t.add(nums[l]); result.add(t); k++; l--; while(k<l &&nums[l]==nums[l+1] ){ l--; } while(k<l &&nums[k]==nums[k-1]){ k++; } } } } } return result; } |
You can use two hashtables to get the problem to n^2 complexity.
Not really, because that would require potentially n operations on each loop, making this O(n^4) instead of O(n^3). HashSet avoids this by hashing the input and is O(1).
May be I didn’t copy it correctly, but searching for for-sums of 0 through
an array 0 to 99 gives actually a list of solutions – which is impossible:
Experiment (invoked in Scala):
val a = (0 to 100) toArray
val list = ClassicSolution.fourSum(a, 0)
//this outputs:
//list : java.util.List[java.util.List[Integer]] = [[0, 1, 2, 100], [0, 1, 3,
//| 99], [0, 1, 4, 98], [0, 1, 5, 97], [0, 1, 6, 96], [0, 1, 7, 95], [0, 1, 8,
//| 94], [0, 1, 9, 93], [0, 1, 10, 92], [0, 1, 11, 91], [0, 1, 12, 90], [0, 1, 1
//| 3, 89], [0, 1, 14, 88], [0, 1, 15, 87], [0, 1, 16, 86], [0, 1, 17, 85], [0,
//| 1, 18, 84], [0, 1, 19, 83], [0, 1, 20, 82], [0, 1, 21, 81], [0, 1, 22, 80],
//| [0, 1, 23, 79], [0, 1, 24, 78], [0, 1, 25, 77], [0, 1, 26, 76], [0, 1, 27, 7
//| 5], [0, 1, 28, 74], [0, 1, 29, 73], [0, 1, 30, 72], [0, 1, 31, 71], [0, 1, 3
//| 2, 70], [0, 1, 33, 69], [0, 1, 34, 68], [0, 1, 35, 67], [0, 1, 36, 66], [0,
//| 1, 37, 65], [0, 1, 38, 64], [0, 1, 39, 63], [0, 1, 40, 62], [0, 1, 41, 61],
//| [0, 1, 42, 60], [0, 1, 43, 59], [0, 1, 44, 58], [0, 1, 45, 57], [0, 1, 46, 5
//| 6], [0, 1, 47, 55], [0, 1, 48, 54], [0, 1, 49, 53], [0, 1, 50, 52], [0, 2, 3
//| , 100], [0, 2, 4, 99], [0, 2, 5, 98], [0, 2, 6, 97], [0, 2, 7, 96], [0, 2, 8
//| , 95], [0, 2, 9, 94], [0
Copied solution:
public class ClassicSolution {
/**
* @param target - what the four integers
* should sum-up together to(typically 0)
*/
public static List<List> fourSum(int[] nums, int target){
List<List> result = new ArrayList();
//if there is no array or if it is smaller then 4
if(nums == null || nums.length < 4){
return result;
}
Arrays.sort(nums);
//outer loop
for (int i = 0; i < nums.length-3; i++) {
/*
* if current int is same as previous
* skip the iteration
*/
if(i != 0 && nums[i] == nums[i-1]) continue;
//first inner loop
for(int j=i+1; j<nums.length-2; j++){
/*
* if not the first iteration or
* f current int is the same as previous
* skip the iteration
*/
if(j != i+1 && nums[j] == nums[j-1])
continue;
int k = j+1; //current inner loop index + 1
int l = nums.length-1; //last index of the passed-in array
/*
* main logic starts here:
* increment k until it hits the end of the array
* remember when we finish we will repeat everything with
* k+1 on next iteration
*/
while(k
* increment k
*/
if(nums[i]+nums[j]+nums[k]+nums[l] target){
//if greater then target - decrement l
l--;
} else{
/*
* save numbers at current indices
* to a new list -> result list
*/
List t = new ArrayList();
t.addAll(Arrays.asList(nums[i],nums[j]
,nums[k],nums[l]));
result.add(t);
k++;
l--;
while(k<l && nums[l] == nums[l+1]){
l--;
}
while(k<l && nums[k] == nums[k-1]){
k++;
}
}
}
}
}
return result;
}
}
This is not a comparable solution due to the extra memory requirement.
I think you don’t need the Set to avoid duplicates if you check for each idx that the current value is different from the last one used. (same logic used in 3Sum solution)
agreed…hashset is not required here
Solution for any sum in 9 lines of Python: https://github.com/renzon/code_interview_training/blob/master/sum_in_array.py
You could simply check if the temp already exist by using result.contains(temp) instead of using HashSet.
For checking duplicate can we do this ?
whenever we find the match get the sum of the i+j+k+l and store it in arraylist and everytime we find a match we can check if that sum exists in the arraylist if not its a new match or else already found.
why not give it a try?
The array is sorted at the beginning.
Good idea!
Shouldn’t it be TreeSet or something instead of ArrayList. The order does matter while comparing one ArrayList to another.
One style optimization of your code is that you don’t need to use both hashset and result.
You just need the hashset.
Set<List> hashset = new HashSet();
When you add tuple of 4, do this:
hashset.add(Arrays.asLists(a, b, c, d)); // pay attention to the desc order
Then, when you return, do this:
return new ArrayList<List>(hashset);
This makes your code clearer
I guess that is because if the duplicate check is based on `while` and `index`, it will require 4 check points every loop. Since there is a `set` to make user there is no duplicate, so no need to check with `while`.
Very nice solution! Can you explain why dont you use the same approach as in 3Sum, i.e. to skip repeating numbers with a while check?
There is a quadratic algorithm for this: http://stackoverflow.com/questions/14732277/quadratic-algorithm-for-4-sum (read first comment). I suppose that you can solve K-sum in O(N^((K+1)/2)) using the “store (K+1)/2-sums in a hash and then search for a match in the hash” logic.