Question

I’m struggling to adapt the code found by codaddict proposed here to solve
this simple variant of the problem. Anyone has an idea?

Answer 1

You take his algorithm, and return high instead of -1. If X[high] is smaller than your target, take the next item instead. If high is equal to your array size, there is no such index.

Michael Anderson is right about how the algorithm you pointed to was written, but it is easy to adapt.

int find_hi (const std::vector<int> &X, int t) {
   int low  = 0;
   int high = X.size() - 1;
   while(low <= high) {
       int mid = (low + high) / 2;
       if(X[mid] == t) return mid;
       else if(X[mid] < t)
           low = mid + 1;
       else
           high = mid - 1;
   }
   if (high < sz && X[high] < t) ++high;
   return high;
}

However, this is equivalent to the following in C++:

int find_hi (const std::vector<int> &X, int t) {
    return std::lower_bound(X.begin(), X.end(), t) - X.begin();
}