Question

While getting ready for interviews, I decided to code the classic “Find if there are two elements in an array that sum up to a given number” question using iterator logic, so that it can be generalized to other containers than vector.

Here’s my function so far

// Search given container for two elements with given sum. 
// If two such elements exist, return true and the iterators 
// pointing to the elements. 
bool hasElementSum( int sum, const vector<int>& v, vector<int>::iterator& el1, vector<int>::iterator& el2 )
{
    bool ret = false;
    el1 = v.begin();
    el2 = v.end()-1;
    while ( el1 != el2 ) {
        if ( *el1 + *el2 == sum ) return true;
        ++el1;--el2;
    }
    return false;
}

This, of course, doesn’t work, but I couldn’t figure out a way to do it without using the condition while ( el1 >= el2 ). Various sources I looked advise against using omnly equality checking for iterators, to be able to generalize to all types of containers that support iterators.

Thanks!

Answer 1

First of all, your algorithm is wrong unless you’ve somehow determined ahead of time that you only need to look at sums where one item is in the first half of the collection, and the other is in the second half of the collection.

If the input’s not sorted, then @sbi’s answer is about as good as it gets.

With a sorted, random-access input, you can start with the first element, and do a binary search (or interpolation search, etc.) to see if you can find the value that would have to go with that to produce the desired sum. Then you can try the second element, but when you do the binary search (or whatever) use the result from the previous search as the upper limit. Since your first element is larger than the previous one, the matching value to produce the correct sum must be less than or equal to what you found the last time around.