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++ 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 + 1][n + 1]; 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 (a[j] == b[i] && j == 0) dp[i][j] = 1; 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; } |
Output:
3
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