Given two strings s and t, determine if they are both one edit distance apart.

Note:

There are 3 possiblities to satisify one edit distance apart:

1. Insert a character into s to get t
2. Delete a character from s to get t
3. Replace a character of s to get t

Example 1:

Input: s = "ab", t = "acb"

Output: true

Explanation: We can insert 'c' into s to get t.

Example 2:

Input: s = "cab", t = "ad"

Output: false

Explanation: We cannot get t from s by only one step.

Example 3:

Input: s = "1203", t = "1213"

Output: true

Explanation: We can replace '0' with '1' to get t.

这个题目思路就是,比较s跟t的长度, 只有abs(s-t) <= 1 才有可能, 然后判断是否只差一个值. 感觉这个题目应该算easy吧.

1. Constraints
1) size >= 0
2) element可以是letter也能是数字

2. Ideas
scan     T: O(n)    S: O(1)

3. Code

 class Solution:

     def oneEdit(self, s,t):

         m, n = len(s), len(t)

         if abs(m-n) >1 or s ==t: return False

         if m > n: s, t, m, n = t, s, n, m

         for i in range(m):

             if s[i] != t[i]:

                 return s[i:] == t[i+1:] or s[i+1: ] == t[i+1:]  # note s[m:] 并不会报错, 但是s[m]会报错!

         return True