# Check if edit distance between two strings is one

An edit between two strings is one of the following changes.

2. Delete a character
3. Change a character

Given two string s1 and s2, find if s1 can be converted to s2 with exactly one edit. Expected time complexity is O(m+n) where m and n are lengths of two strings.

Examples:

```Input:  s1 = "geeks", s2 = "geek"
Output: yes
Number of edits is 1

Input:  s1 = "geeks", s2 = "geeks"
Output: no
Number of edits is 0

Input:  s1 = "geaks", s2 = "geeks"
Output: yes
Number of edits is 1

Input:  s1 = "peaks", s2 = "geeks"
Output: no
Number of edits is 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to find Edit Distance using Dynamic programming. If distance is 1, then return true, else return false. Time complexity of this solution is O(n2)

An Efficient Solution is to simultaneously traverse both strings and keep track of count of different characters. Below is complete algorithm.

```Let the input strings be s1 and s2 and lengths of input
strings be m and n respectively.

1) If difference between m an n is more than 1,
return false.
2) Initialize count of edits as 0.
3) Start traversing both strings from first character.
a) If current characters don't match, then
(i)   Increment count of edits
(ii)  If count becomes more than 1, return false
(iii) If length of one string is more, then only
possible  edit is to remove a character.
Therefore, move ahead in larger string.
(iv)  If length is same, then only possible edit
is to  change a character. Therefore, move
b) Else, move ahead in both strings. ```

Below is the implementation of the above idea :

## C++

 `// C++ program to check if given two strings are ` `// at distance one. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns true if edit distance between s1 and ` `// s2 is one, else false ` `bool` `isEditDistanceOne(string s1, string s2) ` `{ ` `    ``// Find lengths of given strings ` `    ``int` `m = s1.length(), n = s2.length(); ` ` `  `    ``// If difference between lengths is more than ` `    ``// 1, then strings can't be at one distance ` `    ``if` `(``abs``(m - n) > 1) ` `        ``return` `false``; ` ` `  `    ``int` `count = 0; ``// Count of edits ` ` `  `    ``int` `i = 0, j = 0; ` `    ``while` `(i < m && j < n) ` `    ``{ ` `        ``// If current characters don't match ` `        ``if` `(s1[i] != s2[j]) ` `        ``{ ` `            ``if` `(count == 1) ` `                ``return` `false``; ` ` `  `            ``// If length of one string is ` `            ``// more, then only possible edit ` `            ``// is to remove a character ` `            ``if` `(m > n) ` `                ``i++; ` `            ``else` `if` `(m< n) ` `                ``j++; ` `            ``else` `//Iflengths of both strings is same ` `            ``{ ` `                ``i++; ` `                ``j++; ` `            ``} ` `             `  `            ``// Increment count of edits  ` `            ``count++; ` `        ``} ` ` `  `        ``else` `// If current characters match ` `        ``{ ` `            ``i++; ` `            ``j++; ` `        ``} ` `    ``} ` ` `  `    ``// If last character is extra in any string ` `    ``if` `(i < m || j < n) ` `        ``count++; ` ` `  `    ``return` `count == 1; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `   ``string s1 = ``"gfg"``; ` `   ``string s2 = ``"gf"``; ` `   ``isEditDistanceOne(s1, s2)? ` `           ``cout << ``"Yes"``: cout << ``"No"``; ` `   ``return` `0; ` `} `

## Java

 `// Java program to check if given  ` `// two strings are at distance one. ` `class` `GFG  ` `{ ` `// Returns true if edit distance  ` `// between s1 and s2 is one, else false ` `static` `boolean` `isEditDistanceOne(String s1, ` `                                 ``String s2) ` `{ ` `    ``// Find lengths of given strings ` `    ``int` `m = s1.length(), n = s2.length(); ` ` `  `    ``// If difference between lengths is  ` `    ``// more than 1, then strings can't  ` `    ``// be at one distance ` `    ``if` `(Math.abs(m - n) > ``1``) ` `        ``return` `false``; ` ` `  `    ``int` `count = ``0``; ``// Count of edits ` ` `  `    ``int` `i = ``0``, j = ``0``; ` `    ``while` `(i < m && j < n) ` `    ``{ ` `        ``// If current characters don't match ` `        ``if` `(s1.charAt(i) != s2.charAt(j)) ` `        ``{ ` `            ``if` `(count == ``1``) ` `                ``return` `false``; ` ` `  `            ``// If length of one string is ` `            ``// more, then only possible edit ` `            ``// is to remove a character ` `            ``if` `(m > n) ` `                ``i++; ` `            ``else` `if` `(m< n) ` `                ``j++; ` `            ``else` `// Iflengths of both strings ` `                ``// is same ` `            ``{ ` `                ``i++; ` `                ``j++; ` `            ``} ` `             `  `            ``// Increment count of edits  ` `            ``count++; ` `        ``} ` ` `  `        ``else` `// If current characters match ` `        ``{ ` `            ``i++; ` `            ``j++; ` `        ``} ` `    ``} ` ` `  `    ``// If last character is extra  ` `    ``// in any string ` `    ``if` `(i < m || j < n) ` `        ``count++; ` ` `  `    ``return` `count == ``1``; ` `} ` ` `  `// driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``String s1 = ``"gfg"``; ` `    ``String s2 = ``"gf"``; ` `    ``if``(isEditDistanceOne(s1, s2)) ` `        ``System.out.print(``"Yes"``); ` `    ``else` `        ``System.out.print(``"No"``);  ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python

 `# Python program to check if given two strings are ` `# at distance one ` ` `  `# Returns true if edit distance between s1 and s2 is ` `# one, else false ` `def` `isEditDistanceOne(s1, s2): ` ` `  `    ``# Find lengths of given strings ` `    ``m ``=` `len``(s1) ` `    ``n ``=` `len``(s2) ` ` `  `    ``# If difference between lengths is more than 1, ` `    ``# then strings can't be at one distance ` `    ``if` `abs``(m ``-` `n) > ``1``: ` `        ``return` `false ` ` `  `    ``count ``=` `0`    `# Count of isEditDistanceOne ` ` `  `    ``i ``=` `0` `    ``j ``=` `0` `    ``while` `i < m ``and` `j < n: ` `        ``# If current characters dont match ` `        ``if` `s1[i] !``=` `s2[j]: ` `            ``if` `count ``=``=` `1``: ` `                ``return` `false ` ` `  `            ``# If length of one string is ` `            ``# more, then only possible edit ` `            ``# is to remove a character ` `            ``if` `m > n: ` `                ``i``+``=``1` `            ``elif` `m < n: ` `                ``j``+``=``1` `            ``else``:    ``# If lengths of both strings is same ` `                ``i``+``=``1` `                ``j``+``=``1` ` `  `            ``# Increment count of edits ` `            ``count``+``=``1` ` `  `        ``else``:    ``# if current characters match ` `            ``i``+``=``1` `            ``j``+``=``1` ` `  `    ``# if last character is extra in any string ` `    ``if` `i < m ``or` `j < n: ` `        ``count``+``=``1` ` `  `    ``return` `count ``=``=` `1` ` `  `# Driver program ` `s1 ``=` `"gfg"` `s2 ``=` `"gf"` `if` `isEditDistanceOne(s1, s2): ` `    ``print` `"Yes"` `else``: ` `    ``print` `"No"` ` `  `# This code is contributed by Bhavya Jain `

## C#

 `// C# program to check if given  ` `// two strings are at distance one. ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Returns true if edit distance  ` `// between s1 and s2 is one, else false ` `static` `bool` `isEditDistanceOne(String s1, ` `                              ``String s2) ` `{ ` `     `  `    ``// Find lengths of given strings ` `    ``int` `m = s1.Length, n = s2.Length; ` ` `  `    ``// If difference between lengths is  ` `    ``// more than 1, then strings can't  ` `    ``// be at one distance ` `    ``if` `(Math.Abs(m - n) > 1) ` `        ``return` `false``; ` ` `  `        ``// Count of edits ` `        ``int` `count = 0; ` `        ``int` `i = 0, j = 0; ` `         `  `    ``while` `(i < m && j < n) ` `    ``{ ` `        ``// If current characters  ` `        ``// don't match ` `        ``if` `(s1[i] != s2[j]) ` `        ``{ ` `            ``if` `(count == 1) ` `                ``return` `false``; ` ` `  `            ``// If length of one string is ` `            ``// more, then only possible edit ` `            ``// is to remove a character ` `            ``if` `(m > n) ` `                ``i++; ` `            ``else` `if` `(m< n) ` `                ``j++; ` `                 `  `             ``// If lengths of both  ` `             ``// strings is same ` `            ``else` `            ``{ ` `                ``i++; ` `                ``j++; ` `            ``} ` `             `  `            ``// Increment count of edits  ` `            ``count++; ` `        ``} ` ` `  `        ``// If current characters match ` `        ``else`  `        ``{ ` `            ``i++; ` `            ``j++; ` `        ``} ` `    ``} ` ` `  `    ``// If last character is extra  ` `    ``// in any string ` `    ``if` `(i < m || j < n) ` `        ``count++; ` ` `  `    ``return` `count == 1; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main () ` `{ ` `    ``String s1 = ``"gfg"``; ` `    ``String s2 = ``"gf"``; ` `    ``if``(isEditDistanceOne(s1, s2)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` `} ` `} ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` 1) ` `        ``return` `false; ` ` `  `    ``// Count of edits ` `    ``\$count` `= 0;  ` ` `  `    ``\$i` `= 0; ``\$j` `= 0; ` `    ``while` `(``\$i` `< ``\$m` `&& ``\$j` `< ``\$n``) ` `    ``{ ` `        ``// If current characters ` `        ``// don't match ` `        ``if` `(``\$s1``[``\$i``] != ``\$s2``[``\$j``]) ` `        ``{ ` `            ``if` `(``\$count` `== 1) ` `                ``return` `false; ` ` `  `            ``// If length of one string is ` `            ``// more, then only possible edit ` `            ``// is to remove a character ` `            ``if` `(``\$m` `> ``\$n``) ` `                ``\$i``++; ` `            ``else` `if` `(``\$m``< ``\$n``) ` `                ``\$j``++; ` `                 `  `             ``// If lengths of both  ` `             ``// strings is same ` `            ``else` `            ``{ ` `                ``\$i``++; ` `                ``\$j``++; ` `            ``} ` `             `  `            ``// Increment count of edits  ` `            ``\$count``++; ` `        ``} ` ` `  `        ``// If current characters ` `        ``// match ` `        ``else`  `        ``{ ` `            ``\$i``++; ` `            ``\$j``++; ` `        ``} ` `    ``} ` ` `  `    ``// If last character is ` `    ``// extra in any string ` `    ``if` `(``\$i` `< ``\$m` `|| ``\$j` `< ``\$n``) ` `        ``\$count``++; ` ` `  `    ``return` `\$count` `== 1; ` `} ` ` `  `// Driver Code ` `\$s1` `= ``"gfg"``; ` `\$s2` `= ``"gf"``; ` `if``(isEditDistanceOne(``\$s1``, ``\$s2``)) ` `    ``echo` `"Yes"``; ` `else` `    ``echo` `"No"``; ` ` `  `// This code is contributed by nitin mittal. ` `?> `

Output:

`Yes`

Time complexity: O(n)
Auxiliary Space: O(1)

Thanks to Gaurav Ahirwar for suggesting above solution.

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Improved By : Sam007, nitin mittal

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