# Edit Distance | DP-5

Given two strings str1 and str2 and below operations that can performed on str1. Find minimum number of edits (operations) required to convert ‘str1’ into ‘str2’.

1. Insert
2. Remove
3. Replace

All of the above operations are of equal cost.

Examples:

```Input:   str1 = "geek", str2 = "gesek"
Output:  1
We can convert str1 into str2 by inserting a 's'.

Input:   str1 = "cat", str2 = "cut"
Output:  1
We can convert str1 into str2 by replacing 'a' with 'u'.

Input:   str1 = "sunday", str2 = "saturday"
Output:  3
Last three and first characters are same.  We basically
need to convert "un" to "atur".  This can be done using
below three operations.
Replace 'n' with 'r', insert t, insert a```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

What are the subproblems in this case?
The idea is process all characters one by one staring from either from left or right sides of both strings.
Let us traverse from right corner, there are two possibilities for every pair of character being traversed.

```m: Length of str1 (first string)
n: Length of str2 (second string)
```
1. If last characters of two strings are same, nothing much to do. Ignore last characters and get count for remaining strings. So we recur for lengths m-1 and n-1.
2. Else (If last characters are not same), we consider all operations on ‘str1’, consider all three operations on last character of first string, recursively compute minimum cost for all three operations and take minimum of three values.
1. Insert: Recur for m and n-1
2. Remove: Recur for m-1 and n
3. Replace: Recur for m-1 and n-1

Below is C++ implementation of above Naive recursive solution.

## C++

 `// A Naive recursive C++ program to find minimum number ` `// operations to convert str1 to str2 ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to find minimum of three numbers ` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `{ ` `    ``return` `min(min(x, y), z); ` `} ` ` `  `int` `editDist(string str1, string str2, ``int` `m, ``int` `n) ` `{ ` `    ``// If first string is empty, the only option is to ` `    ``// insert all characters of second string into first ` `    ``if` `(m == 0) ` `        ``return` `n; ` ` `  `    ``// If second string is empty, the only option is to ` `    ``// remove all characters of first string ` `    ``if` `(n == 0) ` `        ``return` `m; ` ` `  `    ``// If last characters of two strings are same, nothing ` `    ``// much to do. Ignore last characters and get count for ` `    ``// remaining strings. ` `    ``if` `(str1[m - 1] == str2[n - 1]) ` `        ``return` `editDist(str1, str2, m - 1, n - 1); ` ` `  `    ``// If last characters are not same, consider all three ` `    ``// operations on last character of first string, recursively ` `    ``// compute minimum cost for all three operations and take ` `    ``// minimum of three values. ` `    ``return` `1 + min(editDist(str1, str2, m, n - 1), ``// Insert ` `                   ``editDist(str1, str2, m - 1, n), ``// Remove ` `                   ``editDist(str1, str2, m - 1, n - 1) ``// Replace ` `                   ``); ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``// your code goes here ` `    ``string str1 = ``"sunday"``; ` `    ``string str2 = ``"saturday"``; ` ` `  `    ``cout << editDist(str1, str2, str1.length(), str2.length()); ` ` `  `    ``return` `0; ` `} `

## Java

 `// A Naive recursive Java program to find minimum number ` `// operations to convert str1 to str2 ` `class` `EDIST { ` `    ``static` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `    ``{ ` `        ``if` `(x <= y && x <= z) ` `            ``return` `x; ` `        ``if` `(y <= x && y <= z) ` `            ``return` `y; ` `        ``else` `            ``return` `z; ` `    ``} ` ` `  `    ``static` `int` `editDist(String str1, String str2, ``int` `m, ``int` `n) ` `    ``{ ` `        ``// If first string is empty, the only option is to ` `        ``// insert all characters of second string into first ` `        ``if` `(m == ``0``) ` `            ``return` `n; ` ` `  `        ``// If second string is empty, the only option is to ` `        ``// remove all characters of first string ` `        ``if` `(n == ``0``) ` `            ``return` `m; ` ` `  `        ``// If last characters of two strings are same, nothing ` `        ``// much to do. Ignore last characters and get count for ` `        ``// remaining strings. ` `        ``if` `(str1.charAt(m - ``1``) == str2.charAt(n - ``1``)) ` `            ``return` `editDist(str1, str2, m - ``1``, n - ``1``); ` ` `  `        ``// If last characters are not same, consider all three ` `        ``// operations on last character of first string, recursively ` `        ``// compute minimum cost for all three operations and take ` `        ``// minimum of three values. ` `        ``return` `1` `+ min(editDist(str1, str2, m, n - ``1``), ``// Insert ` `                       ``editDist(str1, str2, m - ``1``, n), ``// Remove ` `                       ``editDist(str1, str2, m - ``1``, n - ``1``) ``// Replace ` `                       ``); ` `    ``} ` ` `  `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``String str1 = ``"sunday"``; ` `        ``String str2 = ``"saturday"``; ` ` `  `        ``System.out.println(editDist(str1, str2, str1.length(), str2.length())); ` `    ``} ` `} ` `/*This code is contributed by Rajat Mishra*/`

## Python

 `# A Naive recursive Python program to fin minimum number ` `# operations to convert str1 to str2 ` `def` `editDistance(str1, str2, m, n): ` ` `  `    ``# If first string is empty, the only option is to ` `    ``# insert all characters of second string into first ` `    ``if` `m ``=``=` `0``: ` `         ``return` `n ` ` `  `    ``# If second string is empty, the only option is to ` `    ``# remove all characters of first string ` `    ``if` `n ``=``=` `0``: ` `        ``return` `m ` ` `  `    ``# If last characters of two strings are same, nothing ` `    ``# much to do. Ignore last characters and get count for ` `    ``# remaining strings. ` `    ``if` `str1[m``-``1``]``=``=` `str2[n``-``1``]: ` `        ``return` `editDistance(str1, str2, m``-``1``, n``-``1``) ` ` `  `    ``# If last characters are not same, consider all three ` `    ``# operations on last character of first string, recursively ` `    ``# compute minimum cost for all three operations and take ` `    ``# minimum of three values. ` `    ``return` `1` `+` `min``(editDistance(str1, str2, m, n``-``1``),    ``# Insert ` `                   ``editDistance(str1, str2, m``-``1``, n),    ``# Remove ` `                   ``editDistance(str1, str2, m``-``1``, n``-``1``)    ``# Replace ` `                   ``) ` ` `  `# Driver program to test the above function ` `str1 ``=` `"sunday"` `str2 ``=` `"saturday"` `print` `editDistance(str1, str2, ``len``(str1), ``len``(str2)) ` ` `  `# This code is contributed by Bhavya Jain `

## C#

 `// A Naive recursive C# program to ` `// find minimum numberoperations ` `// to convert str1 to str2 ` `using` `System; ` ` `  `class` `GFG { ` `    ``static` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `    ``{ ` `        ``if` `(x <= y && x <= z) ` `            ``return` `x; ` `        ``if` `(y <= x && y <= z) ` `            ``return` `y; ` `        ``else` `            ``return` `z; ` `    ``} ` ` `  `    ``static` `int` `editDist(String str1, String str2, ``int` `m, ``int` `n) ` `    ``{ ` `        ``// If first string is empty, the only option is to ` `        ``// insert all characters of second string into first ` `        ``if` `(m == 0) ` `            ``return` `n; ` ` `  `        ``// If second string is empty, the only option is to ` `        ``// remove all characters of first string ` `        ``if` `(n == 0) ` `            ``return` `m; ` ` `  `        ``// If last characters of two strings are same, nothing ` `        ``// much to do. Ignore last characters and get count for ` `        ``// remaining strings. ` `        ``if` `(str1[m - 1] == str2[n - 1]) ` `            ``return` `editDist(str1, str2, m - 1, n - 1); ` ` `  `        ``// If last characters are not same, consider all three ` `        ``// operations on last character of first string, recursively ` `        ``// compute minimum cost for all three operations and take ` `        ``// minimum of three values. ` `        ``return` `1 + min(editDist(str1, str2, m, n - 1), ``// Insert ` `                       ``editDist(str1, str2, m - 1, n), ``// Remove ` `                       ``editDist(str1, str2, m - 1, n - 1) ``// Replace ` `                       ``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``String str1 = ``"sunday"``; ` `        ``String str2 = ``"saturday"``; ` `        ``Console.WriteLine(editDist(str1, str2, str1.Length, ` `                                   ``str2.Length)); ` `    ``} ` `} ` ` `  `// This Code is Contributed by Sam007 `

## PHP

 ` `

Output:

`3`

The time complexity of above solution is exponential. In worst case, we may end up doing O(3m) operations. The worst case happens when none of characters of two strings match. Below is a recursive call diagram for worst case. We can see that many subproblems are solved, again and again, for example, eD(2, 2) is called three times. Since same suproblems are called again, this problem has Overlapping Subprolems property. So Edit Distance problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array that stores results of subproblems.

## C++

 `// A Dynamic Programming based C++ program to find minimum ` `// number operations to convert str1 to str2 ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to find the minimum of three numbers ` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `{ ` `    ``return` `min(min(x, y), z); ` `} ` ` `  `int` `editDistDP(string str1, string str2, ``int` `m, ``int` `n) ` `{ ` `    ``// Create a table to store results of subproblems ` `    ``int` `dp[m + 1][n + 1]; ` ` `  `    ``// Fill d[][] in bottom up manner ` `    ``for` `(``int` `i = 0; i <= m; i++) { ` `        ``for` `(``int` `j = 0; j <= n; j++) { ` `            ``// If first string is empty, only option is to ` `            ``// insert all characters of second string ` `            ``if` `(i == 0) ` `                ``dp[i][j] = j; ``// Min. operations = j ` ` `  `            ``// If second string is empty, only option is to ` `            ``// remove all characters of second string ` `            ``else` `if` `(j == 0) ` `                ``dp[i][j] = i; ``// Min. operations = i ` ` `  `            ``// If last characters are same, ignore last char ` `            ``// and recur for remaining string ` `            ``else` `if` `(str1[i - 1] == str2[j - 1]) ` `                ``dp[i][j] = dp[i - 1][j - 1]; ` ` `  `            ``// If the last character is different, consider all ` `            ``// possibilities and find the minimum ` `            ``else` `                ``dp[i][j] = 1 + min(dp[i][j - 1], ``// Insert ` `                                   ``dp[i - 1][j], ``// Remove ` `                                   ``dp[i - 1][j - 1]); ``// Replace ` `        ``} ` `    ``} ` ` `  `    ``return` `dp[m][n]; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``// your code goes here ` `    ``string str1 = ``"sunday"``; ` `    ``string str2 = ``"saturday"``; ` ` `  `    ``cout << editDistDP(str1, str2, str1.length(), str2.length()); ` ` `  `    ``return` `0; ` `} `

## Java

 `// A Dynamic Programming based Java program to find minimum ` `// number operations to convert str1 to str2 ` `class` `EDIST { ` `    ``static` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `    ``{ ` `        ``if` `(x <= y && x <= z) ` `            ``return` `x; ` `        ``if` `(y <= x && y <= z) ` `            ``return` `y; ` `        ``else` `            ``return` `z; ` `    ``} ` ` `  `    ``static` `int` `editDistDP(String str1, String str2, ``int` `m, ``int` `n) ` `    ``{ ` `        ``// Create a table to store results of subproblems ` `        ``int` `dp[][] = ``new` `int``[m + ``1``][n + ``1``]; ` ` `  `        ``// Fill d[][] in bottom up manner ` `        ``for` `(``int` `i = ``0``; i <= m; i++) { ` `            ``for` `(``int` `j = ``0``; j <= n; j++) { ` `                ``// If first string is empty, only option is to ` `                ``// insert all characters of second string ` `                ``if` `(i == ``0``) ` `                    ``dp[i][j] = j; ``// Min. operations = j ` ` `  `                ``// If second string is empty, only option is to ` `                ``// remove all characters of second string ` `                ``else` `if` `(j == ``0``) ` `                    ``dp[i][j] = i; ``// Min. operations = i ` ` `  `                ``// If last characters are same, ignore last char ` `                ``// and recur for remaining string ` `                ``else` `if` `(str1.charAt(i - ``1``) == str2.charAt(j - ``1``)) ` `                    ``dp[i][j] = dp[i - ``1``][j - ``1``]; ` ` `  `                ``// If the last character is different, consider all ` `                ``// possibilities and find the minimum ` `                ``else` `                    ``dp[i][j] = ``1` `+ min(dp[i][j - ``1``], ``// Insert ` `                                       ``dp[i - ``1``][j], ``// Remove ` `                                       ``dp[i - ``1``][j - ``1``]); ``// Replace ` `            ``} ` `        ``} ` ` `  `        ``return` `dp[m][n]; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``String str1 = ``"sunday"``; ` `        ``String str2 = ``"saturday"``; ` `        ``System.out.println(editDistDP(str1, str2, str1.length(), str2.length())); ` `    ``} ` `} ``/*This code is contributed by Rajat Mishra*/`

## Python

 `# A Dynamic Programming based Python program for edit ` `# distance problem ` `def` `editDistDP(str1, str2, m, n): ` `    ``# Create a table to store results of subproblems ` `    ``dp ``=` `[[``0` `for` `x ``in` `range``(n ``+` `1``)] ``for` `x ``in` `range``(m ``+` `1``)] ` ` `  `    ``# Fill d[][] in bottom up manner ` `    ``for` `i ``in` `range``(m ``+` `1``): ` `        ``for` `j ``in` `range``(n ``+` `1``): ` ` `  `            ``# If first string is empty, only option is to ` `            ``# insert all characters of second string ` `            ``if` `i ``=``=` `0``: ` `                ``dp[i][j] ``=` `j    ``# Min. operations = j ` ` `  `            ``# If second string is empty, only option is to ` `            ``# remove all characters of second string ` `            ``elif` `j ``=``=` `0``: ` `                ``dp[i][j] ``=` `i    ``# Min. operations = i ` ` `  `            ``# If last characters are same, ignore last char ` `            ``# and recur for remaining string ` `            ``elif` `str1[i``-``1``] ``=``=` `str2[j``-``1``]: ` `                ``dp[i][j] ``=` `dp[i``-``1``][j``-``1``] ` ` `  `            ``# If last character are different, consider all ` `            ``# possibilities and find minimum ` `            ``else``: ` `                ``dp[i][j] ``=` `1` `+` `min``(dp[i][j``-``1``],        ``# Insert ` `                                   ``dp[i``-``1``][j],        ``# Remove ` `                                   ``dp[i``-``1``][j``-``1``])    ``# Replace ` ` `  `    ``return` `dp[m][n] ` ` `  `# Driver program ` `str1 ``=` `"sunday"` `str2 ``=` `"saturday"` ` `  `print``(editDistDP(str1, str2, ``len``(str1), ``len``(str2))) ` `# This code is contributed by Bhavya Jain `

## C#

 `// A Dynamic Programming based ` `// C# program to find minimum ` `// number operations to ` `// convert str1 to str2 ` `using` `System; ` ` `  `class` `GFG { ` `    ``static` `int` `min(``int` `x, ``int` `y, ``int` `z) ` `    ``{ ` `        ``if` `(x <= y && x <= z) ` `            ``return` `x; ` `        ``if` `(y <= x && y <= z) ` `            ``return` `y; ` `        ``else` `            ``return` `z; ` `    ``} ` ` `  `    ``static` `int` `editDistDP(String str1, String str2, ``int` `m, ``int` `n) ` `    ``{ ` `        ``// Create a table to store ` `        ``// results of subproblems ` `        ``int``[, ] dp = ``new` `int``[m + 1, n + 1]; ` ` `  `        ``// Fill d[][] in bottom up manner ` `        ``for` `(``int` `i = 0; i <= m; i++) { ` `            ``for` `(``int` `j = 0; j <= n; j++) { ` `                ``// If first string is empty, only option is to ` `                ``// insert all characters of second string ` `                ``if` `(i == 0) ` ` `  `                    ``// Min. operations = j ` `                    ``dp[i, j] = j; ` ` `  `                ``// If second string is empty, only option is to ` `                ``// remove all characters of second string ` `                ``else` `if` `(j == 0) ` ` `  `                    ``// Min. operations = i ` `                    ``dp[i, j] = i; ` ` `  `                ``// If last characters are same, ignore last char ` `                ``// and recur for remaining string ` `                ``else` `if` `(str1[i - 1] == str2[j - 1]) ` `                    ``dp[i, j] = dp[i - 1, j - 1]; ` ` `  `                ``// If the last character is different, consider all ` `                ``// possibilities and find the minimum ` `                ``else` `                    ``dp[i, j] = 1 + min(dp[i, j - 1], ``// Insert ` `                                       ``dp[i - 1, j], ``// Remove ` `                                       ``dp[i - 1, j - 1]); ``// Replace ` `            ``} ` `        ``} ` ` `  `        ``return` `dp[m, n]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``String str1 = ``"sunday"``; ` `        ``String str2 = ``"saturday"``; ` `        ``Console.Write(editDistDP(str1, str2, str1.Length, ` `                                 ``str2.Length)); ` `    ``} ` `} ` `// This Code is Contributed by Sam007 `

## PHP

 ` `

Output:

`3`

Time Complexity: O(m x n)
Auxiliary Space: O(m x n)

Space Complex Solution: In the above-given method we require O(m x n) space. This will not be suitable if the length of strings is greater than 2000 as it can only create 2D array of 2000 x 2000. To fill a row in DP array we require only one row the upper row. For example, if we are filling the i = 10 rows in DP array we require only values of 9th row. So we simply create a DP array of 2 x str1 length. This approach reduces the space complexity. Here is the C++ implementation of the above-mentioned problem.

 `// A Space efficient Dynamic Programming ` `// based C++ program to find minimum ` `// number operations to convert str1 to str2 ` `#include ` `using` `namespace` `std; ` ` `  `void` `EditDistDP(string str1, string str2) ` `{ ` `    ``int` `len1 = str1.length(); ` `    ``int` `len2 = str2.length(); ` ` `  `    ``// Create a DP array to memoize result ` `    ``// of previous computations ` `    ``int` `DP[len1 + 1]; ` ` `  `    ``// To fill the DP array with 0 ` `    ``memset``(DP, 0, ``sizeof` `DP); ` ` `  `    ``// Base condition when second string ` `    ``// is empty then we remove all characters ` `    ``for` `(``int` `i = 0; i <= len1; i++) ` `        ``DP[i] = i; ` ` `  `    ``// Start filling the DP ` `    ``// This loop run for every ` `    ``// character in second string ` `    ``for` `(``int` `i = 1; i <= len2; i++) { ` `        ``// This loop compares the char from ` `        ``// second string with first string ` `        ``// characters ` `        ``for` `(``int` `j = 0; j <= len1; j++) { ` `            ``// if first string is empty then ` `            ``// we have to perform add character ` `            ``// operation to get second string ` `            ``if` `(j == 0) ` `                ``DP[i % 2][j] = i; ` ` `  `            ``// if character from both string ` `            ``// is same then we do not perform any ` `            ``// operation . here i % 2 is for bound ` `            ``// the row number. ` `            ``else` `if` `(str1[j - 1] == str2[i - 1]) { ` `                ``DP[i % 2][j] = DP[(i - 1) % 2][j - 1]; ` `            ``} ` ` `  `            ``// if character from both string is ` `            ``// not same then we take the minimum ` `            ``// from three specified operation ` `            ``else` `{ ` `                ``DP[i % 2][j] = 1 + min(DP[(i - 1) % 2][j], ` `                                       ``min(DP[i % 2][j - 1], ` `                                           ``DP[(i - 1) % 2][j - 1])); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// after complete fill the DP array ` `    ``// if the len2 is even then we end ` `    ``// up in the 0th row else we end up ` `    ``// in the 1th row so we take len2 % 2 ` `    ``// to get row ` `    ``cout << DP[len2 % 2][len1] << endl; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``string str1 = ``"food"``; ` `    ``string str2 = ``"money"``; ` `    ``EditDistDP(str1, str2); ` `    ``return` `0; ` `} `

Output:

`4`

Time Complexity: O(m x n)
Auxiliary Space: O( m )

Applications: There are many practical applications of edit distance algorithm, refer Lucene API for sample. Another example, display all the words in a dictionary that are near proximity to a given wordincorrectly spelled word.

Thanks to Vivek Kumar for suggesting updates.