import java.util.*;

/** 
 *  Sequential and binary searches
 *  Code from CSE142 lecture, 2/03
 *  Author: Hal Perkins
 *  (Note: to use this code, you would need to add strings to the list and,
 *  for binary search, the strings would have to be in alphabetical order.)
 */
public class Search {
  // instance variable
  private ArrayList names;         // list of Strings
  
  /** Construct empty list */
  public Search() {
    names = new ArrayList();
  }
  
  /** Add str to the list */
  public void add(String str) {
    names.add(str);
  }
  
  /** Return a string with the list elements */
  public String toString() {
    if (names.size() == 0) {
      return "";
    }
    // at least one string
    String result = (String)names.get(0);
    for (int k = 1; k < names.size(); k++) {
      result = result + " " + (String)names.get(k);
    }
    return result;
  }
  
  /** return location of str in the list, or -1 if not found */
  public int linearSearch(String str) {
    // linear (sequential) search
    for (int k = 0; k < names.size(); k++) {
      if (str.equals((String)names.get(k))) {
        return k;
      }
    }
    // not found if control reaches here
    return -1;
  }
  
  /** return location of str in the list, or -1 if not found.
   *  precondition: List is sorted, i.e., names(1) <= names(2) <= ... */
  public int binarySearch(String str) {
    // binary search
    // invariant: names(0..L) <= str, names(R..size()-1) > str
    //            names(L+1..R-1) is the unknown region in the middle
    int L = -1;
    int R = names.size();
    while (L+1 != R) {
      int mid = (L + R) / 2;
      if (((String)names.get(mid)).compareTo(str) <= 0) {
        L = mid;
      } else {
        R = mid;
      }
    }
      
    // if found, str equals names(L), but need to guard against L=-1 if all
    // elements in names are > str
    if (L >= 0 && ((String)names.get(L)).equals(str)) {
      return L;
    } else {
      return -1;
    }
  }
}