[LC] 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:

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.

class Solution {
    public boolean isOneEditDistance(String s, String t) {
        if (Math.abs(s.length() - t.length()) > 1) {
            return false;
        }
        if (s.length() == t.length()) {
            return isOneModify(s, t);
        } else if (s.length() > t.length()) {
            return isOneDel(s, t);
        } else {
            return isOneDel(t, s);
        }
    }

    private boolean isOneModify(String s, String t) {
        int diff = 0, i = 0;
        while (i < s.length()) {
            if (s.charAt(i) != t.charAt(i)) {
                diff += 1;
            }
            i += 1;
        }
        return diff == 1;
    }

    private boolean isOneDel(String s, String t) {
        int i = 0;
        for (; i < s.length() && i < t.length(); i++) {
            if (s.charAt(i) != t.charAt(i)) {
                break;
            }
        }
        return s.substring(i + 1).equals(t.substring(i));
    }
}