Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). count8(8) → 1 co…

Count8.java

public int count8(int n) {
  if(n==0)
    return 0;
  int rem100=n%100;
  
  int div=n/10;
  int rem=n%10;
  if(rem100==88){
    return 2+count8(div);
  }
  if(rem==8){
    return 1+count8(div);
  }
  return count8(div);
}