One Edit Distance in C++

Suppose we have two strings s and t; we have to check whether they are both one edit distance apart. The one edit distance has three types −

• Insert a character into s to get t

• Delete a character from s to get t

• Replace a character of s to get t

So, if the input is like s = "ab", t = "acb", then the output will be True

To solve this, we will follow these steps −

• n := size of s, m := size of t

• if n < m, then −

  • return isOneEditDistance(t, s)

• for initialize i := 0, when i < m, update (increase i by 1), do −

  • if s[i] is not equal to t[i], then −

    • if n is same as m, then −

      • return true when substring of s from index 0 to (i) is the same as the substring of t from index 0 to (i)

    • return true when substring of s from index 0 to (i) is the same as the substring of t from index 0 to (i - 1)

• return true when m + 1 is the same as n

Example 

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool isOneEditDistance(string s, string t) {
      int n = s.size();
      int m = t.size();
      if (n < m) {
         return isOneEditDistance(t, s);
      }
      for (int i = 0; i < m; i++) {
         if (s[i] != t[i]) {
            if (n == m) {
               return s.substr(i + 1) == t.substr(i + 1);
            }
            return s.substr(i + 1) == t.substr(i);
         }
      }
      return m + 1 == n;
   }
};
main(){
   Solution ob;
   cout << (ob.isOneEditDistance("ab", "acb"));
}

Input

s = "ab", t = "acb"

Output

1