**One Edit Distance - Amazon Top Interview Questions**

### Problem Statement :

Given two strings s0 and s1 determine whether they are one or zero edit distance away. An edit can be described as deleting a character, adding a character, or replacing a character with another character. Constraints n ≤ 100,000 where n is the length of s0. m ≤ 100,000 where m is the length of s1. Example 1 Input s0 = "quicksort" s1 = "quicksort" Output True Explanation This has the edit distance of 0, since they are the same string. Example 2 Input s0 = "mergesort" s1 = "mergesorts" Output True Explanation This has the edit distance 1, since s was added to the second string. Example 3 Input s0 = "mergeport" s1 = "mergesorts" Output False Explanation This has edit distance of 2.

### Solution :

` ````
Solution in C++ :
bool equal(string& s0, int i, string& s1, int j) {
while (i < s0.size() && j < s1.size() && s0[i] == s1[j]) {
i++, j++;
}
return i == s0.size() && j == s1.size();
}
bool solve(string s0, string s1) {
int m = s0.size(), n = s1.size();
if (m > 1 + n || n > 1 + m) return false;
for (int i = 0; i < min(m, n); i++) {
if (s0[i] != s1[i])
return equal(s0, i + 1, s1, i) || equal(s0, i, s1, i + 1) ||
equal(s0, i + 1, s1, i + 1);
}
return true;
}
```

` ````
Solution in Java :
class Solution {
public boolean solve(String s0, String s1) {
int m = s0.length(), n = s1.length();
if (Math.abs(m - n) > 1) {
return false;
}
int i = 0, j = 0, edit = 0;
while (i < m && j < n) {
if (s0.charAt(i) != s1.charAt(j)) {
edit += 1;
if (m < n) {
i--;
} else if (m > n) {
j--;
}
}
i++;
j++;
}
return (edit < 2);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s0, s1):
# Get dimensions
n = len(s0)
m = len(s1)
# Ensure n < m
if m < n:
return self.solve(s1, s0)
# Check if length diff is > 1
if abs(n - m) > 1:
return False
# index i -> s0, index j -> s1
i = 0
j = 0
edits = 0
while i < n:
# chars match, increment i, j :D
if s0[i] == s1[j]:
i += 1
j += 1
# mismatch, count edit >:(
else:
edits += 1
# same length, replace char in s0, move i, j
if n == m:
i += 1
j += 1
# not same length, delete char from s1 by moving j
else:
j += 1
# too many edits, terminate
if edits > 1:
break
# 1 edit or less?
return edits <= 1
```

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