# Ways of transforming one string to other by removing 0 or more characters

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

**Examples:**

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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 Explanation : 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 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` `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 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.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 characters 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_` |

## Javascript

`<script>` `// JavaScript program to count the` `// distinct transformation` `// of one string to other.` `function` `countTransformation(a,b)` ` ` `{` ` ` `var` `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;` ` ` `}` ` ` `var` `dp = ` `new` `Array (m,n);` ` ` `// Fil dp[][] in bottom up manner` ` ` `// Traverse all character of b[]` ` ` `for` `(` `var` `i = 0; i < m; i++) {` ` ` `// Traverse all characters of a[] for b[i]` ` ` `for` `(` `var` `j = i; j < n; j++) {` ` ` `// Filling the first row of the dp` ` ` `// matrix.` ` ` `if` `(i == 1) {` ` ` `if` `(j == 1) {` ` ` `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[j]) {` ` ` `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` ` ` `var` `a = ` `"abcccdf"` `, b = ` `"abccdf"` `;` ` ` `document.write(countTransformation(a, b));` `// This code is contributed by shivanisinghss2110` `</script>` |

**Output:**

3

**Time Complexity:** O(n^2)

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