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[LintCode] Edit Distance II

Given two strings S and T, determine if they are both one edit distance apart. There are three types of

edits that can be performed on strings: insert a character, remove a character, or replace a character.

Example

Given s = "aDb", t = "adb"
return true

If s is one insertion apart from t, then t is one deletion apart from s as insertion and deletion are reverse

operations. So we simplify the original problem to given two strings smaller and bigger, with smaller's length

less than or equal to bigger's length, check if smaller is one insertion or replace apart from bigger.

Algorithm:

1. From the start of both strings, compare each character. If equal, increment both index pointers by 1;

2. If not equal: increment the difference count by 1, if this count is > 1, return false;

if the two strings have the same length, then this difference is the only replace that should

　　　　 happen, increment both index pointers and expect no more differences.

if they don't have the same length, then it means this difference is where an insertion could happen

to make them equal. Only increment bigger's index pointer and expect no more differences.

Runtime: O(smaller string's length)

 1 public class Solution {
 2     /**
 3      * @param s a string
 4      * @param t a string
 5      * @return true if they are both one edit distance apart or false
 6      */
 7     public boolean isOneEditDistance(String s, String t) {
 8         if(s == null || t == null || Math.abs(s.length() - t.length()) > 1){
 9             return false;
10         }
11         String smaller = s.length() <= t.length() ? s : t;
12         String bigger = s.length() > t.length() ? s : t;
13         int smallerIdx = 0, biggerIdx = 0;
14         int diffCnt = 0;
15         while(smallerIdx < smaller.length()){
16             if(smaller.charAt(smallerIdx) != bigger.charAt(biggerIdx)){
17                 diffCnt++;
18                 if(diffCnt > 1){
19                     return false;
20                 }
21                 if(smaller.length() == bigger.length()){
22                     smallerIdx++;
23                     biggerIdx++;
24                 }
25                 else{
26                     biggerIdx++;
27                 }
28             }
29             else{
30                 smallerIdx++;
31                 biggerIdx++;
32             }
33         }
34         if(smaller.length() == bigger.length() && diffCnt == 0){
35             return false;
36         }
37         return true;
38     }
39 }