I think you have unnecessarily complicated it, or otherwise, I have misunderstood the problem.
If you want to check if K operations can make the array non positive, consider subtracting B from everything K times. And then you want to check if you can reduce K more elements by (A-B) to make everything non positive. It should be simply O( N \log a[i]_{max} ) .
yep, you are correct
i kinda felt that there is a much simpler solution than binarysearching for x_i(number of times a is removed) but was too lazy to think about it
thanks a lot 
can you please provide sudo code for your approach?
int count = 0;
sort(int a[],int n){
int temp,i,j;
for(i=0;i<n-1;i++){
for(j=i;j<n;j++)
if(a[i]>a[j]){
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
subtract(int a[],int n,int b, int c){
count++;
a[n-1]=a[n-1]-b;
int i;
for(i=0;i<n-1;i++){
a[i]=a[i]-c;
}
}
int main(){
int i,j,n,k=0,b,c;
printf(“Enter size of array :”);
scanf("%d",&n);
int a[n];
printf(“Enter b which is big value”);
scanf("%d",&b);
printf(“Enter c which is small value”);
scanf("%d",&c);
for(i=0;i<n;i++){
printf(“Enter elements :”);
scanf("%d",&a[i]);
}
while(k==0){
k=1;
for(i=0;i<n;i++){
if(a[i]>-1){
k=0;
}
}
sort(a,n);
subtract(a,n,b,c);
}
printf("%d",count-1);
}
Here’s the implementation of smarth gupta’s idea
#include<bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(long i=a;i<b;i++)
#define pb push_back
#define ll long long
#define fastio ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
//================================================//
int main(){
fastio;
ll n,a,b;
cin>>n>>a>>b;
ll A[n];
ll maxi=-9;
FOR(i,0,n){cin>>A[i];maxi=max(maxi,A[i]);}
ll l=0,r=maxi,mid;
ll ansi=1e17;
while(l<=r){
mid=(l+r)/2;
ll B[n];
bool flag=true;
ll bb=mid*b;
ll op=mid;
ll aa=(a-b);
FOR(i,0,n){ B[i]=A[i]; B[i]-=bb;}
FOR(i,0,n){
while(B[i]>0&&op>0){
ll mul=B[i]/aa;
ll rem=B[i]%aa;
if(rem!=0){mul++;}
if(mul<=op)
{ B[i]-=aa*mul;
op-=mul;}else{break;}
}
if(B[i]>0){flag=false; break;}
}
if(flag){ r=mid-1; ansi=min(ansi,mid); }
else{l=mid+1; }
}
cout<<ansi<<"\n";
return 0;}
O(N log a[i] max)
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
ll n;
cin>>n;
vector<ll> g(n);
for(auto&u:g) cin>>u;
ll a,b;
cin>>a>>b;
ll l=0,r=1e9;
ll dif=a-b;
ll ans=r;
while(l<=r)
{
ll m=l+(r-l)/2;
vector<ll> v=g;
for(ll i=0;i<n;i++) v[i]-=(m*b);
ll rem=m;
bool ok=true;
for(ll i=0;i<n;i++)
{
if(v[i]>0 and rem>0)
{
ll req=(v[i]+dif-1)/dif;
ll used=min(rem,req);
v[i]-=(used*dif);
rem-=used;
if(v[i]>0)
{
ok=false;
break;
}
}
}
if(ok)
{
ans=m;
r=m-1;
}
else l=m+1;
}
cout<<ans<<endl;
return 0;
}
I think this is what u meant guptosity sir.
Regards
Mr Trash
@tamo11 Have a look at this one please and if possible can you give me some direction for the solution. Thanks in advance!