4Sum Problem Analysis: O(N^2) VS O(N^2logN) VS O(N^3)


I intended to do a thorough analysis for K sum problems, so here we go for the 4sum problem. The common approach is to reduce it to 3sum and then to 2sum with O(N^3) time, this approach is discussed and implemented. The hash based approach trying to reduce the time complexity is analyzed too and we can achieve different time complexity O(N^2) or O(N^2logN) depending on tree hash map or unordered hash (copyright @sigmainfy) map we use. Again, comments are highly appreciated and that could help the readers and me to find anything we might miss!

4Sum Problem Definition:

Given an array S of n integers, are there elements abc, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.


  • 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)

4Sum Problem Algorithm Analysis

1) Reduce the problem into 3Sum and then 2Sum: this is similar trick we use as in “3Sum Problem Analysis: Handling Duplicates Without Using Set“. That is, we linear scan the sorted array, and when we process the current element A[i], we just do 3sum in the rest of the array starting from i+1, the next would be exactly the same with 3sum problem, same tricks to compare the current value with the previous value is used to avoid or remove the duplicate quadruplets in the final results, and we don’t need to do any filtering using say Set at the end. The sorting would take O(NlogN) and the total time would be O(N^3) which is not difficult to figure out, the following is the implementation and it is accepted by the leetcode OJ:

class Solution {
    vector< vector<int> > fourSum(vector<int> &num, int target) {
		vector<vector<int> > vecQuadruplets;
		if (num.size() < 4) return vecQuadruplets;

		vector<int> quadruplet(4, 0);
		sort(num.begin(), num.end());
		int lastIndex4 = num.size() - 3;
		int lastIndex3 = num.size() - 2;
		for (int i = 0; i < lastIndex4; ++i) {
			quadruplet[0] = num[i];
			if (i > 0 && num[i] == num[i - 1]) continue;
			for (int j = i + 1; j < lastIndex3; ++j) {
				if (j > i + 1 && num[j] == num[j - 1]) 

				quadruplet[1] = num[j];
				int p = j + 1, q = num.size() - 1;
				while (p < q) {
					if (p > j + 1 && num[p] == num[p - 1]) {
					int sumOver = num[i] + num[j] + num[p] + num[q];
					if (sumOver == target) {
						quadruplet[2] = num[p++];
						quadruplet[3] = num[q--];
					else if (sumOver < target) {
					else {
		return vecQuadruplets;

2) Hash approach: here I propose the following hash approach and depending on which data structure you use, the time complexity would be different, could be O(N^2logN) or O(N^2). Here we go:
Basic Idea:

  1. We insert all the pairs formed from any two out of the input into a hashmap, let’s call it HashTable, this step takes O(N^2 * t) where t is the time complexity for the hash insertion operation. I will explain t a bit later.
  2. HashTable is such a table: each entry in the this hash table is like this,  HashTable[hashvalue] = list, where the list is a linked list structure contains all the pairs the sum of which equal to the key value.
  3. So the next thing we want to do is to find the two sum which equal to the given target among all the pair sum values which are just the unique keys in our hashmap HashTable. This involves iterate all the whole hash map.
  4. The last step is to iterate the HashTable, and for each key value say we currently process Key A, we need to check (B = target – A) whether B exist in the hash table or not, if so, then we have potential candidates of two pair of pairs (i.e., a quadruplets), one of the pair picked from list HashTable[A] and the other picked from list HashTable[B].


I give some remarkable things about the above steps and time complexity would be formalized at the end after several details are discussed.

  • We need to keep the original index pair in the list at each key slot of the hasthtable, i.e., HashTable[PairSumValue] = list, the list contains pairs, and pair<int, int> = (a, b) , that is  a and b are original index such that num[a] + num[b] = PairSumValue. This is because we don’t want to use the same element in the original array more than once. For example, in our setting, we might pick two pairs (1, 2) and (2, 3) in step 4 because num[1] + num[2] + num[2] + num[3] = target, if we just store values, we cannot identify this is an invalid  quadruplet, if we store index instead, obviously (1, 2, 2, 3) is not a legal quadruplet in our setting, we only need to make sure (copyright @sigmainfy) all the indexes in quadruplets are different and note we shall return values rather than index which is easy to do conversion. This additional checking takes O(1) time.
  • As I said at the beginning, the type of the hashmap affects the time complexity, in step 4, we actually need to linear scan all the keys, for tree based map which is ordered map it is straightforward, for unordered hash map, we could do this: in step 1, when we make the hashmap, at the same time, we push all the unique pair values into a separate vector or array, that is, each time we got a pair sum which is not in the hashmap yet, we push this pair sum value into the separate vector, at the end of step 1, this vector (let’s call v1) would of size O(N^2), we iterate this vector v1 instead of the unordered hash map and the access to the unordered hash map takes O(1). Otherwise, we will have to scan all the memory space in the unordered hashmap to iterate all the pair sum values, which would take O(M) where M is the size of the unordered hashmap and it could be much larger than N^2.
  • Time complexity: time complexity depends on which kind of hash map we use. With all the details in our mind, If we use unordered hash map, then the value t would be t = O(1) and if we use tree map like red-black tree which is ordered map, then t = O(logN), so for unordered map, step 1 takes O(N^2), step 4 takes O(N^2), and the total time would be O(N^2); For tree map which is ordered, step 1 takes O(N^2logN) because insert takes O(logN) and steps 4 takes O(N^2logN) too because access the key takes O(logN).

Note to really implement the hash approach, I think it is quite sophysicated handling all the details. The time complexity might still be arguable, any comments are welcome and highly appreciated!!!


To summarize, I have the sorting approach to reduce the 4sum problem into 3sum problem and then 2sum, it takes O(N^3) time. And I discussed my way to tackle this problem by using hash and discussed the associated sophysicated details, the time complexity depends on the types of the hash table we use.

Written on September 22, 2014