BinarySearch - One Edit Distance

[https://binarysearch.com/problems/One-Edit-Distance](https://binarysearch.com/problems/One-Edit-Distance) ## Problem 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. ## Solution Short and clean. Find the first index i that s0[i] is not equal to s1[i] Based on the length of s0 and s1, compare the rest of the sub string are same or not. ``` bool solve(string s0, string s1) { int m = s0.size(), n = s1.size(); for (int i = 0; i < min(m, n); i++) { if (s0[i] != s1[i]) { if (m == n) return s0.substr(i + 1) == s1.substr(i + 1); else if (m < n) return s0.substr(i) == s1.substr(i + 1); else return s0.substr(i + 1) == s1.substr(i); } } return abs(m - n) <= 1; } ```