A Plane Journey - HackerEarth ( Binary Search )

A flight company has to schedule a journey of  groups of people from the same source to the same destination. Here, array A represents the number of people in each group. All groups are present at the source. The flight company has  planes where array B represents the capacity of each plane.

You are required to send all groups to destination with the following conditions:

Each plane can travel from the Source to Destination with only one group at a time such that capacity of a plane is enough to accommodate all people in that group.
All people belonging to the same group travel together.
Every plane can make multiple journeys between source and destination.
It costs 1 unit of time to travel between source to destination and vice versa.
Note: Multiple planes can fly together and also it is not necessary for planes to end their journey at the source.

Determine the minimum time required to send all groups from the source to the destination. 

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#include<bits/stdc++.h>
#define ll long long
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
using namespace std;
int main()
{
    IOS;
    ll n,m,cnt=0,i,t,p,j;
    cin>>n>>m;
    vector<ll>a;
    vector<ll>b;
    for(i=0;i<n;i++)
    {
        cin>>p;
        a.push_back(p);
    }
 
    for(i=0;i<m;i++)
    {
        cin>>p;
        b.push_back(p);
    }
 
    sort(a.begin(),a.end());
    sort(b.begin(),b.end());
    if(a[n-1]>b[m-1])
    {
        cout<<"-1";
    }
    else
    {
    int vis[m]={0};
 
    for(i=0;i<n;i++)
    {
        auto it= lower_bound(b.begin(),b.end(),a[i]);
		// it-begin == distance
        if(vis[it-b.begin()]==1)
        {
            while(vis[it-b.begin()]==1)
            {
                it++;
            }
            if(it-b.begin()!=m)
            {
            vis[it-b.begin()]=1;
            cnt++;
            }
        }
        else
        {
            vis[it-b.begin()]=1;
            cnt++;
        }
    }
	// if (n-cnt)==0 means no plane used two times ans will be one considering all planes flied in same time
    cout<<(1 + (n-cnt)*2);
    }
 
 
    return 0;
}