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

Solutions to Chapter 20 | Hard
20.5 You have a large text file containing words. Given any two words, find the shortest
distance (in terms of number of words) between them in the file. Can you make the
searching operation in O(1) time? What about the space complexity for your solution?
SOLUTION
pg 91
We will assume for this question that the word order does not matter. This is a question you
should ask your interviewer. If the word order does matter, we can make the small modification
shown in the code below.
To solve this problem, simply traverse the file and for every occurrence of word1 and word2,
compare difference of positions and update the current minimum.
1 int shortest(String[] words, String word1, String word2) {
2 int pos = 0;
3 int min = Integer.MAX_VALUE / 2;
4 int word1_pos = -min;
5 int word2_pos = -min;
6 for (int i = 0; i < words.length; i++) {
7 String current_word = words[i];
8 if (current_word.equals(word1)) {
9 word1_pos = pos;
10 // Comment following 3 lines if word order matters
11 int distance = word1_pos - word2_pos;
12 if (min > distance)
13 min = distance;
14 } else if (current_word.equals(word2)) {
15 word2_pos = pos;
16 int distance = word2_pos - word1_pos;
17 if (min > distance) min = distance;
18 }
19 ++pos;
20 }
21 return min;
22 }
To solve this problem in less time (but more space), we can create a hash table with each
word and the locations where it occurs. We then just need to find the minimum (arithmetic)
difference in the locations (e.g., abs(word0.loc[1] - word1.loc[5])).
To find the minimum arithmetic difference, we take each location for word1 (e.g.: 0, 3} and
do a modified binary search for it in word2’s location list, returning the closest number. Our
search for 3, for example, in {2, 7, 9} would return 1. The minimum of all these binary searches
is the shortest distance.
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Cracking the Coding Interview | Additional Review Problems