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

Solutions to Chapter 20 | Hard
20.7 Write a program to find the longest word made of other words.
SOLUTION
pg 91
The solution below does the following:
1. Sort the array by size, putting the longest word at the front
2. For each word, split it in all possible ways. That is, for “test”, split it into {“t”, “est”}, {“te”,
“st”} and {“tes”, “t”}.
3. Then, for each pairing, check if the first half and the second both exist elsewhere in the
array.
4. “Short circuit” by returning the first string we find that fits condition #3.
What is the time complexity of this?
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Time to sort array: O(n log n)
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Time to check if first / second half of word exists: O(d) per word, where d is the average
length of a word.
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Total complexity: O(n log n + n * d). Note that d is fixed (probably around 5—10 characters).
Thus, we can guess that for short arrays, the time is estimated by O(n * d) , which
also equals O(number of characters in the array). For longer arrays, the time will be better
estimated by O(n log n).
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Space complexity: O(n).
Optimizations: If we didn’t want to use additional space, we could cut out the hash table. This
would mean:
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Sorting the array in alphabetical order
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Rather than looking up the word in a hash table, we would use binary search in the array
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We would no longer be able to short circuit.
1 class LengthComparator implements Comparator {
2 @Override
3 public int compare(String o1, String o2) {
4 if (o1.length() < o2.length()) return 1;
5 if (o1.length() > o2.length()) return -1;
6 return 0;
7 }
8 }
2 8 7
Cracking the Coding Interview | Additional Review Problems