Given an array of integers and a value, determine if there are any two integers in the array whose sum is equal to the given value. Return true if the sum exists and return false if it does not.

Consider this array and the target sums:

- Use hashing
- Use comparison between elements

bool find_sum_of_two(vector<int>& A, int val) { //TODO: Write - Your - Code return false; }

bool find_sum_of_two(vector<int>& A, int val) { unordered_set<int> found_values; for (int& a : A) { if (found_values.find(val - a) != found_values.end()) { return true; } found_values.insert(a); } return false; } int main() { vector<int> v = {5, 7, 1, 2, 8, 4, 3}; vector<int> test = {3, 20, 1, 2, 7}; for(int i=0; i<test.size(); i++){ bool output = find_sum_of_two(v, test[i]); cout << "find_sum_of_two(v, " << test[i] << ") = " << (output ? "true" : "false") << endl; } return 0; }

The runtime complexity of this solution is linear, O(n).

The memory complexity of this solution is linear, O(n).

In this solution, you can use the following algorithm to find a pair that add up to the target (say `val`

).

- Scan the whole array once and store visited elements in a hash set.
- During scan, for every element
`e`

in the array, we check if`val`

-`e`

is present in the hash set i.e.`val`

-`e`

is already visited.- If
`val`

-`e`

is found in the hash set, it means there is a pair (`e`

,`val`

-`e`

) in array whose sum is equal to the given`val`

. - If we have exhausted all elements in the array and didn’t find any such pair, the function will return
`false`

.

- If