Four sum problem to find tuplets - space complexity?

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I am working on below interview question:

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ? N ? 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.



A = [ 1, 2]

B = [-2,-1]

C = [-1, 2]

D = [ 0, 2]

Output: 2

Below is the code and I am not able to understand why the space complexity is O(n^2)? We use only one Map so it should be O(n) space complexity?

public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
    Map<Integer, Integer> map = new HashMap<>();

    for(int i=0; i<C.length; i++) {
        for(int j=0; j<D.length; j++) {
            int sum = C[i] + D[j];
            map.put(sum, map.getOrDefault(sum, 0) + 1);

    int res=0;
    for(int i=0; i<A.length; i++) {
        for(int j=0; j<B.length; j++) {
            res += map.getOrDefault(-1 * (A[i]+B[j]), 0);

    return res;

answered question

2 Answers


it depends on the length of input arrays A,B,C,D

if we assume each array with length N that means we have ,

O(NxN) for the first nested loop O(NxN) for the second nested loop

hence we have O(NxN) + O(NxN) = O(2xN^2)

= O(N^2)

posted this

Let's say C.length = D.length = N. You have a loop inside loop. It means, in the first part of the algorithm you call map.put function N*N times. It gives you already O(N^2). As you use the HashMap you can assume that get and put operations "cost" constant time.

The second part more or less the same. It gives you O(N^2 + N^2) = O(N^2).

posted this

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