package com.freetymekiyan.algorithms.level.medium;
/**
* This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.
* <p>
* Given a list of words and two words word1 and word2, return the shortest distance between these two words in the
* list.
* <p>
* word1 and word2 may be the same and they represent two individual words in the list.
* <p>
* For example,
* Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
* <p>
* Given word1 = “makes”, word2 = “coding”, return 1.
* Given word1 = "makes", word2 = "makes", return 3.
* <p>
* Note:
* You may assume word1 and word2 are both in the list.
* <p>
* Company Tags: LinkedIn
* Tags: Array
* Similar Problems: (E) Shortest Word Distance, (M) Shortest Word Distance II
*/
public class ShortestWordDistance3 {
/**
* Array. Only one index.
* Remember the previous index word1 or word2 seen.
* Iterate through the array.
* If we find either word1 or word2.
* If previous index is initialized,
* and the two words are the same or previous word is not the same as the current one.
* Update min.
* Then we update previous index.
*/
public int shortestWordDistance(String[] words, String word1, String word2) {
int prevIndex = -1;
int min = words.length;
boolean same = word1.equals(word2);
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1) || words[i].equals(word2)) {
if (prevIndex != -1 && (same || !words[prevIndex].equals(words[i]))) {
min = Math.min(i - prevIndex, min);
}
prevIndex = i;
}
}
return min;
}
/**
* Array. Two indices.
* Words may be the same but indices are not.
* i1 and i2 are the last seen positions for word1 and word2.
* When word1 and word2 are the same,
* i1 is the first position, i2 is the latest position.
*/
public int shortestWordDistanceB(String[] words, String word1, String word2) {
int i1 = -1;
int i2 = -1;
int shortest = Integer.MAX_VALUE;
boolean same = word1.equals(word2);
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
if (same) {
i1 = i2;
i2 = i;
} else {
i1 = i;
}
} else if (words[i].equals(word2)) {
i2 = i;
}
shortest = Math.min(shortest, Math.abs(i1 - i2));
}
return shortest;
}
}