[LC] 161. One Edit Distance
2024-10-08 20:49:43
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
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));
}
}
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