Program 6TH in C++

PROBLEM STATEMENT :

Find two no’s in a sorted array whose sum =X ?
i gave a solution whose complexity is O(N) .. Anyone with reduced time complexity?

SOLUTION:

#include <iostream.h>

#include <cstdio>

#include <list.h>

#include <unistd.h>

#include <vector.h>

using namespace std;

int main()

{

int val=0;

int sum=0;

int flag=1;

int diff=0;

int i=0,j;

list<int> num1;

//Entering Array

cout<<"\n\nEnter numbers for the array :\n\n";

while(true)

{

cout<<"\n\nEnter : ";

cin>>val;

if(cin.good())

{

num1.push_back(val);

continue;

}

else

{

cout<<"\n\nYou entered a non integral value \n\n";

cin.clear();

cin.ignore(numeric_limits<streamsize>::max(),'\n');

cin.sync();

break;

}

}

num1.sort();

vector<int> num (num1.begin(),num1.end());

num1.clear();

//Entering the sum user wants to find

cout<<"\n\nEnter the sum you want to find : ";

while(true)

{

cout<<"\n\nEnter : ";

cin>>sum;

if(cin.good())

{

break;

}

else

{

cout<<"\n\nYou entered a non integral value \n\n";

cin.clear();

cin.ignore(numeric_limits<streamsize>::max(),'\n');

cin.sync();

continue;

}

}

if(num.back()>sum)

{

num.pop_back();

}

j=num.size()-1;

//Required loop to find the numbers if present

while(i < j)

{

diff=sum-num[i];

if(diff < num.back())

{

while(diff < num.back())

{

num.pop_back();

}

j=num.size()-1;

continue;

}

else

if(diff == num.back())

{

flag=0;

break;

}

i++;

}

if(flag == 0)

{

cout<<"\n\nThe numbers are at index : "<<i<<" and "<<num.size()-1;

cout<<"\n\nThe numbers are : "<<num[i]<<" and "<<num.back();

cout<

}

else

{

cout<<"\n\n No such numbers are present int the array \n\n";

}

}