# Number of values in a vector that add up to a sum S

I have the following task:

Write the function that has the header:long long CountSumS(vector &a, int s)

The function will return the number of pairs with (a[i], a[j]); i < j; a[i] + a[j] = s)
Example:
If a = (8, 2, 3, 8, 7, 5) and s = 10, the function will return 3, the pairs being (8, 2), (2, 8) and (3, 7)

I’m also trying to do this efficiently, so simply taking every other element and checking the sum of all the values after it isn’t the solution I’m looking for.

I tried solving this using an unordered multimap and I have the following function:

``````long long CountSumS(vector<int>& v, int sum)
{
int cnt = 0;
typedef unordered_multimap<int, int>::iterator iter;
unordered_multimap<int, int> m;

for (int i = 0; i < v.size(); i++)
m.insert(make_pair(v[i], i));

for (int i = 0; i < v.size(); i++)
{
int x = sum - v[i];
//the needed value to make the sum

pair<iter, iter> result = m.equal_range(x);
int count = distance(result.first, result.second);

if (count != 0)
{
for (auto iter = result.first; iter != result.second; iter ++) //iterates through the values
{
if (iter->second > i)
cnt++;
/*else
iter = m.erase(iter);*/
}
}
}
return cnt;

``````

It solves the problem inefficiently because after all it just checks through all elements with a certain key, so the solution I thought of was to delete all elements in the map that have value smaller than "i", since it means the element has already been passed in the vector and I can’t use it to form a sum anyway. The section written in the /**/ was my approach at trying to erase the value at that iterator, but I get an error when I tried to run the code (image)(https://i.stack.imgur.com/Nl3SI.png). I was wondering what was causing that error.

I appreciate all input on a better aproach for an easier solution, it was just the first thing I could think of and tried to implement.

### >Solution :

Using `iter ++` after `iter = m.erase(iter)` may result in an iterator overrun.

Do only one of them in each iteration.

``````            for (auto iter = result.first; iter != result.second; ) //iterates through the values
{
if (iter->second > i)
{
cnt++;
iter++;
}
else
iter = m.erase(iter);
}
``````