Just think what exactly you want to happen for intervals of size 1, 2 or 3.

You can watch my video here, https://www.youtube.com/watch?v=GU7DpgHINWQ.

Here is what I do:

int start = 0;
int end = n - 1;

while (start <= end) {
    int mid = (start + end) / 2;
    if (check(mid)) LOGIC1
    else LOGIC2
}

The key is to think about what your binary searched values will look like (TTTTTFFFFFF? or opposite) and then about which one you want (first occurrence of T? last occurrence of F?) which mostly depends on the problem and how you defined true/false values. Then you can select either start = mid + 1 or end = mid — 1 in the LOGIC places I wrote. I select these by thinking about what happens at the end case (the destination where we have either TF or FT transition).

I don't think that makes a difference.

PS: I wonder if the second implementation is correct?

What I do is one of 2 of the following, depending on the problem i'm solving:

int lo=begin, hi=end; //the range(inclusive)
while (lo<hi)
{
    int mid=(lo+hi)/2;
    if (f(mid)) lo=mid+1; // can also be !f(mid) if the problem requires that
    else hi=mid;
}
//lo=hi=result

or

int lo=begin, hi=end; //the range(inclusive)
while (lo<hi)
{
    int mid=(lo+hi+1)/2;
    if (f(mid)) lo=mid; // can also be !f(mid) if the problem requires that
    else hi=mid-1;
}
//lo=hi=result

So I start by writing $$$mid=(lo+hi)/2$$$ and when i write the rest of the while loop I add a +1 if I have $$$hi=mid-1$$$.

while(low <= high)
{
   mid = (low + high ) / 2. /// if u use long long .
   else mid = (low + (high &mdash; low)/2) /// if int as it would be safe. 
   do something with mid. 
   store the result of mid if needed.
   then either low = mid + 1 , or high = mid -1 .
}