Given an integer array and a number X, find all unique pairs of (a, b) whose sume is equal to X

Here is an example to the question. Suppose we have an int array = {5, 3, 7, 0, 1, 4, 2} and X = 5. The unique pairs that sum up to 5 are (5, 0) (3, 2) and (1, 4).

There are two approaches to this problem. The first approach is to use brute force which gives time complexity of O(n^2) and space complexity of O(n). We basically just have to loop through the array and for each number we add it to each of other numbers and see if they sum up to 5. If so, we print out the pair. Here is the code for brute force approach:

void findPairOfSum(int arrayOfNum[], int arraySize, int sum)
{
  for (int i = 0; i < arraySize; i++)
  {
    for (int j = i; j < arraySize; j++)
    {
      if (arrayOfNum[i] + arrayOfNum[j] == sum)
        cout << "(" << arrayOfNum[i] << ", " << arrayOfNum[j] << ")" << endl;
    }
  }
} 

The second approach is to use a hash table to store the difference between sum and each of the elements in the array. Then as we loop through the array, check each element against the hash table. If the element presents, then print out the key value pair. For example, if we hash the example array we'll have this hash table:

Key   Value
5        5 - 5 = 0
3        5 - 3 = 2
7        5 - 7 = -2
0        5 - 0 = 5
1        5 - 1 = 4
4        5 - 4 = 1
2        5 - 3 = 2

This approach will have the time complexity of O(n) and space complexity of O(n). Thus, chose your method wisely, depending on your need (speed or space efficiency).