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# Edit Distance | DP using Memoization

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

• Insert
• Remove
• Replace

All of the above operations are of equal cost.

Examples:

Input: str1 = “geek”, str2 = “gesek”
Output:
We can convert str1 into str2 by inserting a ‘s’.

Input: str1 = “cat”, str2 = “cut”
Output:
We can convert str1 into str2 by replacing ‘a’ with ‘u’.

Input: str1 = “sunday”, str2 = “saturday”
Output:
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

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

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.
• Insert: Recur for m and n-1
• Remove: Recur for m-1 and n
• Replace: Recur for m-1 and n-1

Below is the implementation of the above approach:

## 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()));``    ``}``}`

## Python

 `# A Naive recursive Python program to find 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))`

## 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));``    ``}``}`

## Javascript

 `// A Naive recursive Javascript``// program to find minimum number``// operations to convert str1 to str2` `// Utility function to find minimum of three numbers``function` `min(x, y, z)``{``    ``return` `Math.min(Math.min(x, y), z);``}` `function` `editDist(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` `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``// your code goes here``var` `str1 = ``"sunday"``;``var` `str2 = ``"saturday"``;``console.log( editDist(str1, str2, str1.length, str2.length));`

Output

`3`

Space complexity :- O(MN)

The time complexity of above solution is O(3^n) which is exponential. 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 subproblems are called again, this problem has Overlapping Subproblems 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. The bottom-up approach can be found here.

This code uses top-down dynamic programming and memoization to solve the problem. The repetitive calls in the recursive code can be avoided by using a 2D array to store the results of previous calculations, reducing the time complexity. The 2D array is necessary because two parameters in the recursive calls change value and there are cases of repetitive calls.

Below are the steps:

• Initialize a 2-D DP array of size m *n with -1 at all the index.
• On every recursive call, store the return value at dp[m][n] so that if func(m, n) is called again, it can be answered in O(1) without using recursion.
• Check if the recursive call has been visited previously or not by checking the value at dp[m][n].

Below is the implementation of the above approach:

## C++

 `// A memoization program to find minimum number``// operations to convert str1 to str2``#include ``using` `namespace` `std;` `// Maximum 2-D array column size``const` `int` `maximum = 1000;` `// 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, ``int` `dp[][maximum])``{``    ``// 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 the recursive call has been``    ``// called previously, then return``    ``// the stored value that was calculated``    ``// previously``    ``if` `(dp[m - 1][n - 1] != -1)``        ``return` `dp[m - 1][n - 1];` `    ``// If last characters of two strings are same, nothing``    ``// much to do. Ignore last characters and get count for``    ``// remaining strings.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``if` `(str1[m - 1] == str2[n - 1])``        ``return` `dp[m - 1][n - 1] = editDist(str1, str2, m - 1, n - 1, dp);` `    ``// 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.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``return` `dp[m - 1][n - 1] = 1 + min(editDist(str1, str2, m, n - 1, dp), ``// Insert``                                      ``editDist(str1, str2, m - 1, n, dp), ``// Remove``                                      ``editDist(str1, str2, m - 1, n - 1, dp) ``// Replace``                                      ``);``}` `// Driver Code``int` `main()``{` `    ``string str1 = ``"sunday"``;``    ``string str2 = ``"saturday"``;``    ``int` `m = str1.length();``    ``int` `n = str2.length();` `    ``// Declare a dp array which stores``    ``// the answer to recursive calls``    ``int` `dp[m][maximum];` `    ``// initially all index with -1``    ``memset``(dp, -1, ``sizeof` `dp);` `    ``// Function call``    ``// memoization and top-down approach``    ``cout << editDist(str1, str2, m, n, dp);` `    ``return` `0;``}`

## Java

 `// A memoization program to find minimum number``// operations to convert str1 to str2``import` `java.util.*;` `class` `GFG``{``  ` `  ``// Maximum 2-D array column size``  ``static` `int` `maximum = ``1000``;` `  ``// Utility function to find minimum of three numbers``  ``static` `int` `min(``int` `x, ``int` `y, ``int` `z)``  ``{``    ``return` `Math.min((Math.min(x, y)), z);``  ``}` `  ``static` `int` `editDist(String str1, String str2, ``int` `m,``                      ``int` `n, ``int``[][] dp)``  ``{``    ` `    ``// 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 the recursive call has been``    ``// called previously, then return``    ``// the stored value that was calculated``    ``// previously``    ``if` `(dp[m - ``1``][n - ``1``] != -``1``)``      ``return` `dp[m - ``1``][n - ``1``];` `    ``// If last characters of two strings are same,``    ``// nothing much to do. Ignore last characters and``    ``// get count for remaining strings.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``if` `(str1.charAt(m - ``1``) == str2.charAt(n - ``1``))``      ``return` `dp[m - ``1``][n - ``1``]``      ``= editDist(str1, str2, m - ``1``, n - ``1``, dp);` `    ``// 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.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``return` `dp[m - ``1``][n - ``1``]``      ``= ``1``      ``+ min(editDist(str1, str2, m, n - ``1``,``                     ``dp), ``// Insert``            ``editDist(str1, str2, m - ``1``, n,``                     ``dp), ``// Remove``            ``editDist(str1, str2, m - ``1``, n - ``1``,``                     ``dp) ``// Replace``           ``);``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``String str1 = ``"sunday"``;``    ``String str2 = ``"saturday"``;``    ``int` `m = str1.length();``    ``int` `n = str2.length();` `    ``// Declare a dp array which stores``    ``// the answer to recursive calls``    ``int``[][] dp = ``new` `int``[m][maximum];``    ``for` `(``int` `i = ``0``; i < dp.length; i++)``      ``Arrays.fill(dp[i], -``1``);` `    ``// Function call``    ``// memoization and top-down approach``    ``System.out.println(editDist(str1, str2, m, n, dp));``  ``}``}` `// This code is contributed by Karandeep Singh`

## Python3

 `# A memoization program to find minimum number``# operations to convert str1 to str2``def` `editDistance(str1, str2, m, n, d ``=` `{}):``    ` `    ``key ``=` `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` `key ``in` `d:``        ``return` `d[key]``        ` `    ``# 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.``    ` `    ``# Store the returned value at dp[m-1][n-1]``    ``# considering 1-based indexing``    ``d[key] ``=` `1` `+` `min``(editDistance(str1, str2, m, n ``-` `1``), ``# Insert``                     ``editDistance(str1, str2, m ``-` `1``, n), ``# Remove``                     ``editDistance(str1, str2, m ``-` `1``, n ``-` `1``)) ``# Replace``    ``return` `d[key]` `# Driver code``str1 ``=` `"sunday"``str2 ``=` `"saturday"` `print``(editDistance(str1, str2, ``len``(str1), ``len``(str2)))` `# This code is contributed by puranjanprithu`

## C#

 `// C# code for the above approach``using` `System;` `class` `GFG``{``  ``// Maximum 2-D array column size``  ``static` `int` `maximum = 1000;` `  ``// Utility function to find minimum of three numbers``  ``static` `int` `min(``int` `x, ``int` `y, ``int` `z)``  ``{``    ``return` `Math.Min(Math.Min(x, y), z);``  ``}` `  ``static` `int` `editDist(``string` `str1, ``string` `str2, ``int` `m,``                      ``int` `n, ``int``[,] dp)``  ``{` `    ``// 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 the recursive call has been``    ``// called previously, then return``    ``// the stored value that was calculated``    ``// previously``    ``if` `(dp[m - 1, n - 1] != -1)``      ``return` `dp[m - 1, n - 1];` `    ``// If last characters of two strings are same,``    ``// nothing much to do. Ignore last characters and``    ``// get count for remaining strings.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``if` `(str1[m - 1] == str2[n - 1])``      ``return` `dp[m - 1, n - 1] = editDist(str1, str2, m - 1, n - 1, dp);` `    ``// 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.` `    ``// Store the returned value at dp[m-1][n-1]``    ``// considering 1-based indexing``    ``return` `dp[m - 1, n - 1] = 1 + min(editDist(str1, str2, m, n - 1,``                                               ``dp), ``// Insert``                                      ``editDist(str1, str2, m - 1, n,``                                               ``dp), ``// Remove``                                      ``editDist(str1, str2, m - 1, n - 1,``                                               ``dp) ``// Replace``                                     ``);``  ``}` `  ``// Driver code``  ``static` `void` `Main(``string``[] args)``  ``{``    ``string` `str1 = ``"sunday"``;``    ``string` `str2 = ``"saturday"``;``    ``int` `m = str1.Length;``    ``int` `n = str2.Length;` `    ``// Declare a dp array which stores``    ``// the answer to recursive calls``    ``int``[,] dp = ``new` `int``[m, maximum];``    ``for` `(``int` `i = 0; i < dp.GetLength(0); i++)``      ``for` `(``int` `j = 0; j < dp.GetLength(1); j++)``        ``dp[i, j] = -1;` `    ``// Function call``    ``// memoization and top-down approach``    ``Console.WriteLine(editDist(str1, str2, m, n, dp));``  ``}``}` `// This code is contributed by lokeshpotta20.`

## Javascript

 `// A memoization function to find the minimum number of operations to convert str1 to str2``function` `editDist(str1, str2, m, n, dp) {``  ``// If the first string is empty, the only option is to``  ``// insert all characters of the second string into the first``  ``if` `(m === 0) ``return` `n;` `  ``// If the second string is empty, the only option is to``  ``// remove all characters of the first string``  ``if` `(n === 0) ``return` `m;` `  ``// if the recursive call has been``  ``// called previously, then return``  ``// the stored value that was calculated``  ``// previously``  ``if` `(dp[m - 1][n - 1] !== undefined) ``return` `dp[m - 1][n - 1];` `  ``// If the last characters of two strings are the same, there's not much to do.``  ``// Ignore last characters and get the count for the remaining strings.` `  ``// Store the returned value at dp[m-1][n-1]``  ``// considering 1-based indexing``  ``if` `(str1[m - 1] === str2[n - 1])``    ``return` `(dp[m - 1][n - 1] = editDist(str1, str2, m - 1, n - 1, dp));` `  ``// If the last characters are not the same, consider all three``  ``// operations on the last character of the first string, recursively``  ``// compute the minimum cost for all three operations and take``  ``// the minimum of three values.` `  ``// Store the returned value at dp[m-1][n-1]``  ``// considering 1-based indexing``  ``return` `(dp[m - 1][n - 1] =``    ``1 +``    ``Math.min(``      ``editDist(str1, str2, m, n - 1, dp), ``// Insert``      ``editDist(str1, str2, m - 1, n, dp), ``// Remove``      ``editDist(str1, str2, m - 1, n - 1, dp) ``// Replace``    ``));``}` `// Driver code``function` `main() {``  ``let str1 = ``"sunday"``;``  ``let str2 = ``"saturday"``;``  ``let m = str1.length;``  ``let n = str2.length;` `  ``// Declare a dp array which stores``  ``// the answer to recursive calls``  ``let dp = Array(m)``    ``.fill(0)``    ``.map(() => Array(n).fill(undefined));` `  ``// Function call``  ``// memoization and top-down approach``  ``console.log(editDist(str1, str2, m, n, dp));``}` `main();`

Output

`3`

Complexity Analysis:

• Time Complexity: O(M * N)
• Auxiliary Space: O(M * N)