给定两个序列A，B，找出序列A中许多独特的方式，以形成与序列B相同的A子序列。转换是通过将字符串A(通过删除0个或多个字符)转换为字符串B来进行的。
例子：
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.
询问：谷歌
解决此问题的想法是使用动态编程。构造一个m * n大小的2D DP矩阵，其中m是字符串B的大小，n是字符串A的大小。
dp [i] [j]给出了将字符串A [0…j]转换为B [0…i]的方式的数量。
- 情况1 ：dp [0] [j] = 1，因为将B =“”与A的任何子字符串一起放置将只有一种解决方案，即删除A中的所有字符。
- 情况2：当i> 0时，可以通过两种情况得出dp [i] [j]：
- 情况2.a：如果B [i]！= A [j]，则解决方案是忽略字符A [j]，并将子字符串B [0..i]与A [0 ..(j-1 )]。因此，dp [i] [j] = dp [i] [j-1]。
- 情况2.b：如果B [i] == A [j]，那么首先我们可以得到情况a)的解，但是我们也可以匹配字符B [i]和A [j]并将其余的它们(即B [0 ..(i-1)]和A [0 ..(j-1)]。因此，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. #includeusing 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_
输出：
3时间复杂度： O(n ^ 2)