Given two sequences A, B, find out number of unique ways in sequence A, to form a subsequence of A that is identical to the sequence B. Transformation is meant by converting string A (by removing 0 or more characters) to string B.

**Examples:**

Input : A = "abcccdf", B = "abccdf" Output : 3 Explanation : Three ways will be -> "ab.ccdf", "abc.cdf" & "abcc.df" . "." is where character is removed. Input : A = "aabba", B = "ab" Output : 4 Expalnation : Four ways will be -> "a.b..", "a..b.", ".ab.." & ".a.b." . "." is where characters are removed.

Asked in : Google

The idea to solve this problem is using Dynamic Programming. Construct a 2D DP matrix of m*n size, where m is size of string B and n is size of string A.**dp[i][j] **gives the number of ways of transforming string A[0…j] to B[0…i].

**Case 1**: dp[0][j] = 1, since placing B = “” with any substring of A would have only 1 solution which is to delete all characters in A.**Case 2 :**when i > 0, dp[i][j] can be derived by two cases:**Case 2.a :**if B[i] != A[j], then the solution would be to ignore the character A[j] and align substring B[0..i] with A[0..(j-1)]. Therefore, dp[i][j] = dp[i][j-1].**Case 2.b :**if B[i] == A[j], then first we could have the solution in case a), but also we could match the characters B[i] and A[j] and place the rest of them (i.e. B[0..(i-1)] and A[0..(j-1)]. As a result, dp[i][j] = dp[i][j-1] + dp[i-1][j-1].

## C++

`// C++ program to count the distinct transformation`

`// of one string to other.`

`#include <bits/stdc++.h>`

`using`

`namespace`

`std;`

`int`

`countTransformation(string a, string b)`

`{`

`int`

`n = a.size(), m = b.size();`

`// If b = "" i.e., an empty string. There`

`// is only one way to transform (remove all`

`// characters)`

`if`

`(m == 0)`

`return`

`1;`

`int`

`dp[m][n];`

`memset`

`(dp, 0,`

`sizeof`

`(dp));`

`// Fil dp[][] in bottom up manner`

`// Traverse all character of b[]`

`for`

`(`

`int`

`i = 0; i < m; i++) {`

`// Traverse all charaters of a[] for b[i]`

`for`

`(`

`int`

`j = i; j < n; j++) {`

`// Filling the first row of the dp`

`// matrix.`

`if`

`(i == 0) {`

`if`

`(j == 0)`

`dp[i][j] = (a[j] == b[i]) ? 1 : 0;`

`else`

`if`

`(a[j] == b[i])`

`dp[i][j] = dp[i][j - 1] + 1;`

`else`

`dp[i][j] = dp[i][j - 1];`

`}`

`// Filling other rows.`

`else`

`{`

`if`

`(a[j] == b[i])`

`dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1];`

`else`

`dp[i][j] = dp[i][j - 1];`

`}`

`}`

`}`

`return`

`dp[m - 1][n - 1];`

`}`

`// Driver code`

`int`

`main()`

`{`

`string a =`

`"abcccdf"`

`, b =`

`"abccdf"`

`;`

`cout << countTransformation(a, b) << endl;`

`return`

`0;`

`}`

## Java

`// Java program to count the`

`// distinct transformation`

`// of one string to other.`

`class`

`GFG {`

`static`

`int`

`countTransformation(String a,`

`String b)`

`{`

`int`

`n = a.length(), m = b.length();`

`// If b = "" i.e., an empty string. There`

`// is only one way to transform (remove all`

`// characters)`

`if`

`(m ==`

`0`

`) {`

`return`

`1`

`;`

`}`

`int`

`dp[][] =`

`new`

`int`

`[m][n];`

`// Fil dp[][] in bottom up manner`

`// Traverse all character of b[]`

`for`

`(`

`int`

`i =`

`0`

`; i < m; i++) {`

`// Traverse all charaters of a[] for b[i]`

`for`

`(`

`int`

`j = i; j < n; j++) {`

`// Filling the first row of the dp`

`// matrix.`

`if`

`(i ==`

`0`

`) {`

`if`

`(j ==`

`0`

`) {`

`dp[i][j] = (a.charAt(j) == b.charAt(i)) ?`

`1`

`:`

`0`

`;`

`}`

`else`

`if`

`(a.charAt(j) == b.charAt(i)) {`

`dp[i][j] = dp[i][j -`

`1`

`] +`

`1`

`;`

`}`

`else`

`{`

`dp[i][j] = dp[i][j -`

`1`

`];`

`}`

`}`

`// Filling other rows.`

`else`

`if`

`(a.charAt(j) == b.charAt(i)) {`

`dp[i][j] = dp[i][j -`

`1`

`]`

`+ dp[i -`

`1`

`][j -`

`1`

`];`

`}`

`else`

`{`

`dp[i][j] = dp[i][j -`

`1`

`];`

`}`

`}`

`}`

`return`

`dp[m -`

`1`

`][n -`

`1`

`];`

`}`

`// Driver code`

`public`

`static`

`void`

`main(String[] args)`

`{`

`String a =`

`"abcccdf"`

`, b =`

`"abccdf"`

`;`

`System.out.println(countTransformation(a, b));`

`}`

`}`

`// This code is contributed by`

`// PrinciRaj1992`

## Python3

`# Python3 program to count the distinct`

`# transformation of one string to other.`

`def`

`countTransformation(a, b):`

`n`

`=`

`len`

`(a)`

`m`

`=`

`len`

`(b)`

`# If b = "" i.e., an empty string. There`

`# is only one way to transform (remove all`

`# characters)`

`if`

`m`

`=`

`=`

`0`

`:`

`return`

`1`

`dp`

`=`

`[[`

`0`

`]`

`*`

`(n)`

`for`

`_`

`in`

`range`

`(m)]`

`# Fill dp[][] in bottom up manner`

`# Traverse all character of b[]`

`for`

`i`

`in`

`range`

`(m):`

`# Traverse all charaters of a[] for b[i]`

`for`

`j`

`in`

`range`

`(i, n):`

`# Filling the first row of the dp`

`# matrix.`

`if`

`i`

`=`

`=`

`0`

`:`

`if`

`j`

`=`

`=`

`0`

`:`

`if`

`a[j]`

`=`

`=`

`b[i]:`

`dp[i][j]`

`=`

`1`

`else`

`:`

`dp[i][j]`

`=`

`0`

`elif`

`a[j]`

`=`

`=`

`b[i]:`

`dp[i][j]`

`=`

`dp[i][j`

`-`

`1`

`]`

`+`

`1`

`else`

`:`

`dp[i][j]`

`=`

`dp[i][j`

`-`

`1`

`]`

`# Filling other rows`

`else`

`:`

`if`

`a[j]`

`=`

`=`

`b[i]:`

`dp[i][j]`

`=`

`(dp[i][j`

`-`

`1`

`]`

`+`

`dp[i`

`-`

`1`

`][j`

`-`

`1`

`])`

`else`

`:`

`dp[i][j]`

`=`

`dp[i][j`

`-`

`1`

`]`

`return`

`dp[m`

`-`

`1`

`][n`

`-`

`1`

`]`

`# Driver Code`

`if`

`__name__`

`=`

`=`

`"__main__"`

`:`

`a`

`=`

`"abcccdf"`

`b`

`=`

`"abccdf"`

`print`

`(countTransformation(a, b))`

`# This code is contributed by vibhu4agarwal`

## C#

`// C# program to count the distinct transformation`

`// of one string to other.`

`using`

`System;`

`class`

`GFG {`

`static`

`int`

`countTransformation(`

`string`

`a,`

`string`

`b)`

`{`

`int`

`n = a.Length, m = b.Length;`

`// If b = "" i.e., an empty string. There`

`// is only one way to transform (remove all`

`// characters)`

`if`

`(m == 0)`

`return`

`1;`

`int`

`[, ] dp =`

`new`

`int`

`[m, n];`

`for`

`(`

`int`

`i = 0; i < m; i++)`

`for`

`(`

`int`

`j = 0; j < n; j++)`

`dp[i, j] = 0;`

`// Fil dp[][] in bottom up manner`

`// Traverse all character of b[]`

`for`

`(`

`int`

`i = 0; i < m; i++) {`

`// Traverse all characters of a[] for b[i]`

`for`

`(`

`int`

`j = i; j < n; j++) {`

`// Filling the first row of the dp`

`// matrix.`

`if`

`(i == 0) {`

`if`

`(j == 0)`

`dp[i, j] = (a[j] == b[i]) ? 1 : 0;`

`else`

`if`

`(a[j] == b[i])`

`dp[i, j] = dp[i, j - 1] + 1;`

`else`

`dp[i, j] = dp[i, j - 1];`

`}`

`// Filling other rows.`

`else`

`{`

`if`

`(a[j] == b[i])`

`dp[i, j] = dp[i, j - 1] + dp[i - 1, j - 1];`

`else`

`dp[i, j] = dp[i, j - 1];`

`}`

`}`

`}`

`return`

`dp[m - 1, n - 1];`

`}`

`// Driver code`

`static`

`void`

`Main()`

`{`

`string`

`a =`

`"abcccdf"`

`, b =`

`"abccdf"`

`;`

`Console.Write(countTransformation(a, b));`

`}`

`}`

`// This code is contributed by DrRoot_`

**Output:**3

**Time Complexity:**O(n^2)This article is contributed by

**Jatin Goyal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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