c++ - Algorithm to count the number of pairs whose sum exists in the array -

given array of n integers. how can count number of pairs sum exists in array?

e.g. int a[] = {1, 3, 4, 6, 7}. here there 3 such pairs: (1 + 3) = 4, (3 + 4) = 7, (1+6) = 7.

there no duplicate numbers in given array , array not sorted.also array can changed,its not necessary maintain array.

i have tried following 2 codes want reduce complexity of code less o(n^2).

try 1:(complexity o(n^3))

#include<iostream> #include<vector> #include<algorithm> using namespace std; main(){     int i,input,j,n,ans=0,sum;     cin>>n;     vector<int> vec;     for(i=0;i<n;i++){         cin>>input;         vec.push_back(input);     }     for(i=0;i<n;i++){         for(j=i+1;j<n;j++){             sum=vec[i]+vec[j];             if(find(vec.begin(),vec.end(),sum)!= vec.end()){                 ans++;             }         }     }     cout<<ans; } 

i'm not sure if reduce complexity, think it's simpler way since don't declare variables in function you're using, saving memory , time:

#include<iostream> #include<vector> #include<algorithm> using namespace std; bool checkforeveryelement(vector<int> vec,int sum) {     for( int i=0; i<vec.size(); i++)     {         for(int j=i+1; j<vec.size(); j++)            if(vec[i] + vec[j] == sum)              return 1;     }         return 0; } void main(){     int i,input,j,n,ans=0,sum;     cin>>n;     vector<int> vec;     for(i=0;i<n;i++){         cin>>input;         vec.push_back(input);     }     sort(vec.begin(),vec.end());     for(i=0;i<n;i++)         if(checkforeveryelement(vec,vec[i])){             ans++;         }     cout<<ans; }