# Clean recursive and iterative solutions

• I start with the recursive solution as it is easier for me to think. The idea is find the first character that not equal to each other in s and t, we can either replace that character or delete the one in t. `if(m > n) return isOneEditDistance(t, s);` makes sure that s has less than or equal length to t.

Since it is a tail recursion, it is very easy to change to an iterative solution. I used a for loop to find the first character that doesn't match and compare the substring for possible replace or deletion.

``````// recursive
public class Solution {
public boolean isOneEditDistance(String s, String t) {
int m = s.length();
int n = t.length();
if(m > n) {
return isOneEditDistance(t, s);
}
if(n > m + 1) {
return false;
}
if(m == 0) {
return n == 1;
}
if(s.charAt(0) != t.charAt(0)) {
return s.substring(1).equals(t.substring(1)) || s.equals(t.substring(1));
}
return isOneEditDistance(s.substring(1), t.substring(1));
}
}

// iterative
public class Solution {
public boolean isOneEditDistance(String s, String t) {
int m = s.length();
int n = t.length();
if(m > n) {
return isOneEditDistance(t, s);
}
if(n > m + 1) {
return false;
}
for(int i = 0; i < m; 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 m == n - 1;
}
}``````

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