- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
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 −
#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
To Continue Learning Please Login
Login with Google