# 161. One Edit Distance

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:

Insert a character into s to get t
Delete a character from s to get t
Replace a character of s to get t
<pre>
Example 1:

Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.
</pre>
<pre>
Example 2:

Input: s = "cab", t = "ad"
Output: false
Explanation: We cannot get t from s by only one step.
</pre>
<pre>
Example 3:

Input: s = "1203", t = "1213"
Output: true
Explanation: We can replace '0' with '1' to get t.
</pre>

<details>
<summary>Source code</summary>

**Method 1**
* Naive

class Solution {
public:
    bool isOneEditDistance(string s, string t) {
        if (s.size() < t.size()) return isOneEditDistance(t, s);
        if (s.size() - t.size() >= 2) return false;
        
        bool used = false;
        if (s.size() == t.size()) {
            for (int i = 0; i < s.size(); i++) {
                if (s[i] == t[i]) {
                    continue;
                } else {
                    used = !used;
                    if (!used) {
                        return false;
                    }
                }
            }
        }
        else {
            for (int i = 0, j = 0; i < s.size(); i++, j++) {
                if (s[i] == t[j]) {
                    continue;
                } else {
                    used = !used;
                    if (!used) {
                        return false;
                    }
                    j--;
                }
            }
        }
        
        return (used == true);
    }
};

**Method 2**
* Use substr

class Solution {
public:
    bool isOneEditDistance(string s, string t) {
        if (s.size() < t.size()) return isOneEditDistance(t, s);
        if (s.size() - t.size() >= 2) return false;
        
        for (int i = 0; i < s.size(); i++) {
            if (i >= t.size()) return true;
            if (s[i] == t[i]) continue;
            
            return (s.size() > t.size()) ? (s.substr(i + 1) == t.substr(i))
                                         : (s.substr(i + 1) == t.substr(i + 1));
        }
        
        return false;
    }
};

</details>