An edit between two strings is one of the following changes.
- Add a character
- Delete a character
- 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
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
ahead in both strings.
b) Else, move ahead in both strings.
Below is the implementation of the above idea :
// C++ program to check if given two strings are // at distance one. #include <bits/stdc++.h> 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 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 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# 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. |
<script> // Javascript program to check if given
// two strings are at distance one.
// Returns true if edit distance
// between s1 and s2 is one, else false
function isEditDistanceOne(s1, s2)
{
// Find lengths of given strings
let 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
let count = 0;
let 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;
}
let s1 = "gfg" ;
let s2 = "gf" ;
if (isEditDistanceOne(s1, s2))
document.write( "Yes" );
else
document.write( "No" );
// This code is contributed by decode2207.
</script> |
<?php // PHP program to check if given // two strings are at distance one. // Returns true if edit distance // between s1 and s2 is one, else // false function isEditDistanceOne( $s1 , $s2 )
{ // Find lengths of given strings
$m = strlen ( $s1 );
$n = strlen ( $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 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. ?> |
Yes
Time complexity: O(n)
Auxiliary Space: O(1)
Approach#2: Using dynamic programming
The approach of this code is to calculate the edit distance between two strings using dynamic programming and then check if it is equal to one or not. If the absolute difference in length of the strings is greater than one, it returns False without calculating the edit distance. Otherwise, it fills a 2D array with the minimum number of operations required to transform one string into another, using the operations of insertion, removal, and replacement. Finally, it returns True if the edit distance is equal to one, and False otherwise.
Algorithm
1. If the absolute difference between the lengths of s1 and s2 is greater than 1, return False.
2. Create a 2D array dp of size (n+1) x (m+1) where n is the length of s1 and m is the length of s2.
3. Initialize the first row and first column of dp with values from 0 to m and 0 to n respectively.
4. Traverse the dp array from (1,1) to (n,m) and check the following conditions:
If s1[i-1] and s2[j-1] are equal, set dp[i][j] to dp[i-1][j-1].
Otherwise, set dp[i][j] to 1 + min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]).
5. Return True if dp[n][m] is 1, else return False.
#include <iostream> #include <string> using namespace std;
// Function to check if the edit distance between two strings is one bool isEditDistanceOne(string s1, string s2) {
int n = s1.length(); // Get the length of the first string
int m = s2.length(); // Get the length of the second string
if ( abs (n - m) > 1) // If the absolute difference in string lengths
// is more than 1, return false
return false ;
int dp[n + 1][m + 1]; // Create a 2D array for dynamic programming
// Initialize the dynamic programming array
for ( int i = 0; i <= n; i++) {
for ( int j = 0; j <= m; j++) {
if (i == 0)
dp[i][j] = j; // Initialize the first row with values from 0 to j
else if (j == 0)
dp[i][j] = i; // Initialize the first column with values from 0 to i
else if (s1[i - 1] == s2[j - 1])
dp[i][j] = dp[i - 1][j - 1]; // Characters match, no operation needed
else
dp[i][j] = 1 + min(dp[i][j - 1],
min(dp[i - 1][j], dp[i - 1][j - 1]));
// Characters don't match, calculate the minimum of insert,
// remove, or replace operations
}
}
return dp[n][m] == 1; // Return true if the edit distance is one
} int main() {
string s1 = "gfg" ;
string s2 = "gf" ;
cout << (isEditDistanceOne(s1, s2) ? "true" : "false" ) << endl;
return 0;
} |
public class EditDistanceOne {
// Function to check if the edit distance between two strings is one
public static boolean isEditDistanceOne(String s1, String s2) {
int n = s1.length(); // Get the length of the first string
int m = s2.length(); // Get the length of the second string
if (Math.abs(n - m) > 1 ) // If the absolute difference in string lengths
// is more than 1, return false
return false ;
int [][] dp = new int [n + 1 ][m + 1 ]; // Create a 2D array for dynamic programming
// Initialize the dynamic programming array
for ( int i = 0 ; i <= n; i++) {
for ( int j = 0 ; j <= m; j++) {
if (i == 0 )
dp[i][j] = j; // Initialize the first row with values from 0 to j
else if (j == 0 )
dp[i][j] = i; // Initialize the first column with values from 0 to i
else if (s1.charAt(i - 1 ) == s2.charAt(j - 1 ))
dp[i][j] = dp[i - 1 ][j - 1 ]; // Characters match, no operation needed
else
dp[i][j] = 1 + Math.min(dp[i][j - 1 ],
Math.min(dp[i - 1 ][j], dp[i - 1 ][j - 1 ]));
// Characters don't match, calculate the minimum of insert,
// remove, or replace operations
}
}
return dp[n][m] == 1 ; // Return true if the edit distance is one
}
public static void main(String[] args) {
String s1 = "gfg" ;
String s2 = "gf" ;
System.out.println(isEditDistanceOne(s1, s2));
}
} |
def is_edit_distance_one(s1, s2):
n = len (s1)
m = len (s2)
if abs (n - m) > 1 :
return False
dp = [[ 0 for j in range (m + 1 )] for i in range (n + 1 )]
for i in range (n + 1 ):
for j in range (m + 1 ):
if i = = 0 :
dp[i][j] = j
elif j = = 0 :
dp[i][j] = i
elif s1[i - 1 ] = = s2[j - 1 ]:
dp[i][j] = dp[i - 1 ][j - 1 ]
else :
dp[i][j] = 1 + min (dp[i][j - 1 ], # Insert
dp[i - 1 ][j], # Remove
dp[i - 1 ][j - 1 ]) # Replace
return dp[n][m] = = 1
s1 = "gfg"
s2 = "gf"
print (is_edit_distance_one(s1, s2))
|
using System;
class GFG
{ // Function to check if the edit distance between two strings is one
static bool IsEditDistanceOne( string s1, string s2)
{
int n = s1.Length; // Get the length of the first string
int m = s2.Length; // Get the length of the second string
// If the absolute difference in string lengths
// is more than 1, return false
if (Math.Abs(n - m) > 1)
{
return false ;
}
// Create a 2D array for dynamic programming
int [,] dp = new int [n + 1, m + 1];
// Initialize the dynamic programming array
for ( int i = 0; i <= n; i++)
{
for ( int j = 0; j <= m; j++)
{
if (i == 0)
{
// Initialize the first row with values from 0 to j
dp[i, j] = j;
}
else if (j == 0)
{
// Initialize the first column with values from 0 to i
dp[i, j] = i;
}
else if (s1[i - 1] == s2[j - 1])
{
// Characters match, no operation needed
dp[i, j] = dp[i - 1, j - 1];
}
else
{
// Characters don't match, calculate the minimum of insert,
// remove, or replace operations
dp[i, j] = 1 + Math.Min(dp[i, j - 1], Math.Min(dp[i - 1, j], dp[i - 1, j - 1]));
}
}
}
// Return true if the edit distance is one
return dp[n, m] == 1;
}
static void Main()
{
string s1 = "gfg" ;
string s2 = "gf" ;
Console.WriteLine(IsEditDistanceOne(s1, s2) ? "true" : "false" );
}
} |
// This function checks if the edit distance between two strings is one function is_edit_distance_one(s1, s2) {
let n = s1.length;
let m = s2.length;
// If the absolute difference between the lengths of the strings is greater than 1, edit distance cannot be one
if (Math.abs(n - m) > 1) {
return false ;
}
// Create a 2D array for dynamic programming
let dp = Array.from(Array(n + 1), () => new Array(m + 1).fill(0));
// Fill up the dp array
for (let i = 0; i <= n; i++) {
for (let j = 0; j <= m; j++) {
if (i == 0) {
// If s1 is empty, insert all characters of s2 to s1
dp[i][j] = j;
} else if (j == 0) {
// If s2 is empty, remove all characters of s1
dp[i][j] = i;
} else if (s1[i - 1] == s2[j - 1]) {
// If the characters are same, no operation is needed
dp[i][j] = dp[i - 1][j - 1];
} else {
// If the characters are different, we need to perform one of the three operations - insert, remove or replace
dp[i][j] = 1 + Math.min(dp[i][j - 1], // Insert
dp[i - 1][j], // Remove
dp[i - 1][j - 1]); // Replace
}
}
}
// If edit distance is 1, return true else false
return dp[n][m] == 1;
} // Example usage let s1 = "gfg" ;
let s2 = "gf" ;
console.log(is_edit_distance_one(s1, s2)); |
True
Time complexity: O(n^2)
Auxiliary Space: O(n^2)