Monday, August 24, 2015

Leetcode: Shortest Word Distance

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
Given word1 = “coding”word2 = “practice”, return 3.
Given word1 = "makes"word2 = "coding", return 1.
Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.
Understand the problem:
The problem can be solved by one-pass of the array. 
Iterate through the array, use two pointers pointing to the index of the word1 and word2, maintain the minimum distance. 

Code (Java):
public class Solution {
    public int shortestDistance(String[] words, String word1, String word2) {
        int posA = -1;
        int posB = -1;
        
        int minDistance = Integer.MAX_VALUE;
        
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                posA = i;
            }
            
            if (words[i].equals(word2)) {
                posB = i;
            }
            
            if (posA != -1 && posB != -1) {
                minDistance = Math.min(minDistance, Math.abs(posA - posB));
            }
        }
        
        return minDistance;
    }
}

1 comment:

  1. Instead of this block:
    if (posA != -1 && posB != -1) {
    minDistance = Math.min(minDistance, Math.abs(posA - posB));
    }

    Efficient one is (we donot need to comapre for every word after first occurence):
    if (posA != -1 && posB != -1) {
    minDistance = Math.min(minDistance, Math.abs(posA - posB));
    if(posA >posB)
    posB =-1;
    else
    posA=-1;
    }

    ReplyDelete