This problem requires a greedy approach; scan through the consecutive sets of houses of width $R$, count the number of cameras already there, and if we don't have enough, add as many as needed as far right (towards the next sets of houses) as possible. It is easy to see that adding cameras as far right as possible maximizes their utility for subsequent sets.

To keep the running count, we maintain a trailing and a forward pointer and calculate it incrementally; a quadratic solution will not run in time.

int trail = 0;
int lead = 0;
int sum = 0;
while (lead < R) // calculate number of cameras in initial set
  if (hascamera[lead++])
    sum++;
int r = 0;
while (true) {
  for (int k = lead - 1; sum < 2; k--) // add cameras for this gap
    if (!hascamera[k]) {
      hascamera[k] = true;
      r++;
    }
  if (lead >= N) // done?
    break;
  if (hascamera[trail++]) // advance pointers
    sum--;
  if (hascamera[lead++])
    sum++;
}