checking if 2 numbers of array add up to I

Question

I saw a interview question as follows: Give an unsorted array of integers A and and an integer I, find out if any two members of A add up to I.

any clues?

ti

Answer 1

1. sort the array
2. for each element X in A, perform a binary search for I-X. If I-X is in A, we have a solution.

This is O(nlogn).

If A contains integers in a given (small enough) range, we can use a trick to make it O(n):

1. we have an array V. For each element X in A, we increment V[X].
2. when we increment V[X] we also check if V[I-X] is >0. If it is, we have a solution.

Answer 2

For example, loop and add possible number to set or hash and if found, just return it.

>>> A = [11,3,2,9,12,15]
>>> I = 14
>>> S = set()
>>> for x in A:
...     if x in S:
...         print I-x, x
...     S.add(I-x)
...
11 3
2 12
>>>

Answer 3

O(n) time and O(1) space

If the array is sorted there is a solution in O(n) time complexity.

Suppose are array is array = {0, 1, 3, 5, 8, 10, 14}

And our x1 + x2 = k = 13, so output should be= 5, 8

1. Take two pointers one at start of array, one at end of array
2. Add both the elements at ptr1 and ptr2 array[ptr1] + array[ptr2]
3. if sum > k then decrement ptr2 else increment ptr1
4. Repeat step2 and step3 till ptr1 != ptr2

Same thing explained in detail here. Seems like an Amazon interview Question http://inder-gnu.blogspot.com/2007/10/find-two-nos-in-array-whose-sum-x.html