Find two elements in an array that sum to k

Question

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Given two arrays a and b .Find all pairs of elements (a1,b1) such that a1 belongs to Array A and b1 belongs to Array B whose sum a1+b1 = k .

Given : An unsorted array A of integers
Input : An integer k

Output : All the two element set with sum of elements in each set equal to k in O(n).

Example:

A = {3,4,5,1,4,2}

Input : 6
Output : {3,3}, {5,1}, {4,2}

Note : I know an O(n logn) solution but that would require to have the array sorted. Is there any way by which this problem can be solved in O(n). An non-trivial C++ data structure can be used i.e there's no bound on space

Solution 1

There are k pairs of integers that sum to k: {0,k}, {1,k-1}, ... etc. Create an array B of size k+1 where elements are boolean. For each element e of the array A, if e <= k && B[e] == false, set B[e] = true and if B[k-e] == true, emit the pair {e,k-e}. Needs to be extended slightly for negative integers.

Solution 2

One can reduce the,

Element Uniqueness bit,

to this. No O(n).

Solution 3

Just a simple algorithm off the top of my head:

• Create a bitfield that represents the numbers from 0 to k, labeled B
• For each number i in A
  • Set B[i]
  • If B[k-i] is set, add (i, k-i) to the output

Now as people have raised, if you need to have two instances of the number 3 to output (3, 3) then you just switch the order of the last two statements in the above algorithm.

Also I'm sure that there's a name for this algorithm, or at least some better one, so if anyone knows I'd be appreciative of a comment.

Solution 4

http://codepad.org/QR9ptUwR

This will print all pairs. The algorithm is same as told by @bdares above.

I have used stl maps as we dont have hash tables in STL.

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Comments

[Sanjeevakumar Hiremath]

how did you derive output pair {3,3}? there is only one 3 in input array.

[Admin]

Your algorithm takes time O(k*n), where k is the target number. This is much, much worse than O(n).

[matth]

std::map<>::insert(x) is O(log N), and std::map<>::count(x) is O(N). Isn't your algorithm thus O(N^2)?

[Vink]

@Rob Adams, I guess, count in map is O(logN) as it does not hold multiple elements with same key. Count in multimap is O(N). My algorithm above is O(NlogN). If we use a hashmap instead of map, we would have O(N) algorithm.

[Dominik Grabiec]

No, it's O(n), where n is determined by the length of A. The bitfield is a constant time lookup O(1) not a linear time lookup O(n) which you are thinking.

[Admin]

Yes, you're right. I misread your algorithm, and assumed you were reading every slot in B.

[Chiffa]

As of C++11, we kind of do: unordered_map

[Timofey]

why not to post the code here ;-)

[Timofey]

This algo needs O(k) space and equivalent to the proposed (java) hashset solutions (in c++ set is a tree)

[Dominik Grabiec]

I think you might be confused Tim, I don't use a set data structure. I use a k sized bitfield, so that's 1 bit for each number from 0 to k, hence it uses O( ceil( k / 8 ) ) bytes if you want to be pedantic.

[puneet]

The implementation ideone.com/jOOZX4

[kap]

Can you give some hint what you mean by that? Or a link to a definition of element uniqueness bit and to the reduction?