# LeetCode – Shortest Word Distance III (Java)

This is a follow-up problem of Shortest Word Distance. The only difference is now word1 could be the same as word2.

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

For example,

Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “makes”, word2 = “coding”, return 1.

Given word1 = "makes", word2 = "makes", return 3.

**Java Solution**

In this problem, word1 and word2 can be the same. The two variables used to track indices should take turns to update.

public int shortestWordDistance(String[] words, String word1, String word2) { if(words==null || words.length<1 || word1==null || word2==null) return 0; int m=-1; int n=-1; int min = Integer.MAX_VALUE; int turn=0; if(word1.equals(word2)) turn = 1; for(int i=0; i<words.length; i++){ String s = words[i]; if(word1.equals(s) && (turn ==1 || turn==0)){ m = i; if(turn==1) turn=2; if(n!=-1) min = Math.min(min, m-n); }else if(word2.equals(s) && (turn==2 || turn==0)){ n = i; if(turn==2) turn =1; if(m!=-1) min = Math.min(min, n-m); } } return min; } |

Or we can simply separate two cases (word1 and word2 are the same and not the same):

public int shortestWordDistance(String[] words, String word1, String word2) { if(words==null||words.length==0) return -1; if(word1==null || word2==null) return -1; boolean isSame = false; if(word1.equals(word2)) isSame = true; int shortest= Integer.MAX_VALUE; int prev=-1; int i1=-1; int i2=-1; for(int i=0; i<words.length; i++){ if(isSame){ if(words[i].equals(word1)){ if(prev!=-1){ shortest=Math.min(shortest, i-prev); } prev = i; } }else{ if(word1.equals(words[i])){ i1=i; if(i2!=-1){ shortest = Math.min(shortest, i-i2); } }else if(word2.equals(words[i])){ i2=i; if(i1!=-1){ shortest = Math.min(shortest, i-i1); } } } } return shortest; } |

<pre><code> String foo = "bar"; </code></pre>