CareerCup

first sort the array in nlgn and then its easy just take two pointers one pointing ats start index of array and one at the end of the array and start checking sum
like this
whwrw a is the integer array givcen
int *p;
int *q;
p=a;
q=a+n-1;
while(p!=q)
{
if(*p+*q==sum)
{printf("(%d,%d)\n",*p,*q);
p++;
q--;
continue;}
if(*p+*q<sum)
{
p++;
continue;
}
q--;
}

find all S-A[i], and stick them in a hash table. Check all A[i] against it.

Can't we do it in O(n) without extra memory?

first we should sort d array and calculate sum-a[i] and do binary search for the difference. total runtime takes O(nlogn) i hope

1) Use Hash table<int,int> myMap
Key = Desired Sum - Item In Array
Value = Index
2) For each Item in Array
if not myMap.exist(Arr[i])
Insert in Map <DesiredSum - Arr[i], index>
else
Print Map<DesiredSum - Arr[i]>.Value, Arr[i]

void pairofsums(int *d,int length)
{

int pair1=0,pair2=0;
vector<int>vector1;
map<int,int>possibilites;
for(int x=0;x<length;x++)
{
vector1.push_back(d[x]);
}
for(int i=0;i<vector1.size();i++)
{
cout<<vector1[i]<<endl;
for(int f=0;f<vector1.size();f++)
{
if((vector1[i]+d[f])==2)
{
pair1=vector1[i];
pair2=d[f];
possibilites[pair1]=pair2;
}
}
}
map<int,int>::iterator iter;
for(iter=possibilites.begin();iter!=possibilites.end();iter++)
{
cout<<"Pair 1: "<<iter->first<<" Pair 2: "<<iter->second<<endl;
}
}

what about-5