Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 20 | Hard
20.13 Given a dictionary of millions of words, give an algorithm to find the largest possible
rectangle of letters such that every row forms a word (reading left to right) and every
column forms a word (reading top to bottom).
SOLUTION
pg 92
Many problems involving a dictionary can be solved by doing some preprocessing. Where
can we do preprocessing?
Well, if we’re going to create a rectangle of words, we know that each row must be the same
length and each column must have the same length. So, let’s group the words of the dictionary
based on their sizes. Let’s call this grouping D, where D[i] provides a list of words of
length i.
Next, observe that we’re looking for the largest rectangle. What is the absolute largest rectangle
that could be formed? It’s (length of largest word) * (length of largest word).
1 int max_rectangle = longest_word * longest_word;
2 for z = max_rectangle to 1 {
3 for each pair of numbers (i, j) where i*j = z {
4 /* attempt to make rectangle. return if successful. */
5 }
6 }
By iterating in this order, we ensure that the first rectangle we find will be the largest.
Now, for the hard part: make_rectangle. Our approach is to rearrange words in list1 into
rows and check if the columns are valid words in list2. However, instead of creating, say,
a particular 10x20 rectangle, we check if the columns created after inserting the first two
words are even valid pre-fixes. A trie becomes handy here.
1 WordGroup[] groupList = WordGroup.createWordGroups(list);
2 private int maxWordLength = groupList.length;
3 private Trie trieList[] = new Trie[maxWordLength];
4
5 public Rectangle maxRectangle() {
6 int maxSize = maxWordLength * maxWordLength;
7 for (int z = maxSize; z > 0; z--) {
8 for (int i = 1; i