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

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  • Difficulty Level : Medium
  • Last Updated : 30 Sep, 2022
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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:

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].

Implementation:

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));
 
    // Fill 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];
 
        // Fill 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
                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
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;
 
        // Fill 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);
 
        // Fill 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(m*n)
Auxiliary Space: O(m*n) because it is using extra space for array dp

This article is contributed by Jatin Goyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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