Given two strings s and t, return true if they are both one edit distance apart, otherwise return false.
A string s is said to be one distance apart from a string t if you can:
- Insert exactly one character into s to get t.
- Delete exactly one character from s to get t.
- Replace exactly one character of s with a different character 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 = "", t = ""
Output: false
Explanation: We cannot get t from s by only one step.
Example 3:
Input: s = "a", t = ""
Output: true
Example 4:
Input: s = "", t = "A"
Output: true
Constraints:
- 0 <= s.length <= 10^4
- 0 <= t.length <= 10^4
- s and t consist of lower-case letters, upper-case letters and/or digits.
Solution DFS
class Solution {
public boolean isOneEditDistance(String s, String t) {
return dfs(0, 0, s, t, 1);
}
boolean dfs(int i1, int i2, String s, String t, int editCount) {
if (i1 == s.length() && i2 == t.length()) {
return editCount == 0;
} else if (i1 == s.length() || i2 == t.length()) {
return editCount == 1 && Math.abs(s.length() - t.length()) == 1;
}
if (s.charAt(i1) == t.charAt(i2)) {
return dfs(i1 + 1, i2 + 1, s, t, editCount);
} else {
if (editCount == 0) {
return false;
}
return dfs(i1, i2 + 1, s, t, 0) || dfs(i1 + 1, i2, s, t, 0) || dfs(i1 + 1, i2 + 1, s, t, 0);
}
}
}
Solution Linear
class Solution {
public boolean isOneEditDistance(String s, String t) {
if (s.length() > t.length()) {
return isOneEditDistance(t, s);
}
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != t.charAt(i)) {
return s.substring(i + 1).equals(t.substring(i + 1)) ||
s.substring(i).equals(t.substring(i + 1));
}
}
return Math.abs(s.length() - t.length()) == 1;
}
}