161. One Edit Distance
Explanation:
The problem asks us to determine if two given strings are one edit distance apart. An edit distance is defined as inserting a character, deleting a character, or replacing a character in one of the strings to make them equal.
To solve this problem, we can iterate through both strings simultaneously and compare the characters. If we find a difference, we need to check if it can be resolved within one edit operation. We can handle this by considering three cases:
- If the lengths of the strings differ by more than 1, they can't be one edit distance apart.
- If the lengths are equal, we can allow only one character to be different.
- If the lengths differ by 1, we can allow one string to have an additional character.
Time complexity: O(n) Space complexity: O(1)
:
class Solution {
public boolean isOneEditDistance(String s, String t) {
int m = s.length();
int n = t.length();
if (Math.abs(m - n) > 1) {
return false;
}
for (int i = 0; i < Math.min(m, n); i++) {
if (s.charAt(i) != t.charAt(i)) {
if (m == n) {
return s.substring(i + 1).equals(t.substring(i + 1));
} else if (m < n) {
return s.substring(i).equals(t.substring(i + 1));
} else {
return s.substring(i + 1).equals(t.substring(i));
}
}
}
return Math.abs(m - n) == 1;
}
}Code Editor (Testing phase)
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