/*
 * Copyright (C) 2014 Pedro Vicente Gómez Sánchez.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 * http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package com.github.pedrovgs.problem32;

/**
 * Given two strings passed as argument, can you write a method to return true if the second
 * contains the first one?
 *
 * @author Pedro Vicente Gómez Sánchez.
 */
public class Contains {

  public boolean evaluate(String word1, String word2) {
    return false;
  }
}

@@ -841,16 +841,352 @@
 ains %7B%0A%0A
+  /**%0A   * Iterative solution to this problem. The complexity order of this algorithm is O(N*M) where N%0A   * is the length of the first word and M is the length of the second word. In space terms the%0A   * complexity order of this algorithm is O(1) because we are not using any auxiliary data%0A   * structure to solve this problem.%0A   */%0A
   public
@@ -1211,19 +1211,16 @@
 String w
-ord
 1, Strin
@@ -1226,20 +1226,53 @@
 ng w
-ord2
+2) %7B%0A    if (w1 == null %7C%7C w2 == null
 ) %7B%0A    
 retu
@@ -1271,20 +1271,438 @@
     
+  
-return false
+throw new IllegalArgumentException(%22You can't pass null strings as input.%22);%0A    %7D%0A    boolean contains = false;%0A    for (int i = 0; i %3C w2.length() - 1; i++) %7B%0A      if (w2.charAt(i) == w1.charAt(0)) %7B%0A        for (int j = 0; j %3C w1.length(); j++) %7B%0A          if (w1.charAt(j) == w2.charAt(i + j) && j == w1.length() - 1) %7B%0A            contains = true;%0A            break;%0A          %7D%0A        %7D%0A      %7D%0A    %7D%0A    return contains
 ;%0A  

Add iterative solution to problem 32