Given a two strings S and T, find count of distinct occurrences of T in S as a subsequence.

Examples:

Input : S = banana, T = ban Output : 3 T appears in S as below three subsequences. [ban], [ba n], [b an] Input : S = geeksforgeeks, T = ge Output : 6 T appears in S as below three subsequences. [ge], [ ge], [g e], [g e] [g e] and [ g e]

This problem can be recursively defined as below.

// Returns count of subsequences of S that match T // m is length of T and n is length of S subsequenceCount(S, T, n, m) // An empty string is subsequence of all. 1) If length of T is 0, return 1. // Else no string can be a sequence of empty S. 2) Else if S is empty, return 0. 3) Else if last characters of S and T don't match, remove last character of S and recur for remaining return subsequenceCount(S, T, n-1, m) 4) Else (Last characters match), the result issumof two counts. // Remove last character of S and recur. a) subsequenceCount(S, T, n-1, m) + // Remove last characters of S and T, and recur. b) subsequenceCount(S, T, n-1, m-1)

Since there are overlapping subproblems in above recurrence result, we can apply dynamic programming approach to solve above problem. We create a 2D array mat[m+1][n+1] where m is length of string T and n is length of string S. mat[i][j] denotes the number of distinct subsequence of substring S(1..i) and substring T(1..j) so mat[m][n] contains our solution.

## C++

/* C/C++ program to count number of times S appears as a subsequence in T */ #include <bits/stdc++.h> using namespace std; int findSubsequenceCount(string S, string T) { int m = T.length(), n = S.length(); // T can't appear as a subsequence in S if (m > n) return 0; // mat[i][j] stores the count of occurrences of // T(1..i) in S(1..j). int mat[m + 1][n + 1]; // Initializing first column with all 0s. An empty // string can't have another string as suhsequence for (int i = 1; i <= m; i++) mat[i][0] = 0; // Initializing first row with all 1s. An empty // string is subsequence of all. for (int j = 0; j <= n; j++) mat[0][j] = 1; // Fill mat[][] in bottom up manner for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { // If last characters don't match, then value // is same as the value without last character // in S. if (T[i - 1] != S[j - 1]) mat[i][j] = mat[i][j - 1]; // Else value is obtained considering two cases. // a) All substrings without last character in S // b) All substrings without last characters in // both. else mat[i][j] = mat[i][j - 1] + mat[i - 1][j - 1]; } } /* uncomment this to print matrix mat for (int i = 1; i <= m; i++, cout << endl) for (int j = 1; j <= n; j++) cout << mat[i][j] << " "; */ return mat[m][n] ; } // Driver code to check above method int main() { string T = "ge"; string S = "geeksforgeeks"; cout << findSubsequenceCount(S, T) << endl; return 0; }

## Java

// Java program to count number of times // S appears as a subsequence in T import java.io.*; class GFG { static int findSubsequenceCount(String S, String T) { int m = T.length(); int n = S.length(); // T can't appear as a subsequence in S if (m > n) return 0; // mat[i][j] stores the count of // occurrences of T(1..i) in S(1..j). int mat[][] = new int[m + 1][n + 1]; // Initializing first column with // all 0s. An emptystring can't have // another string as suhsequence for (int i = 1; i <= m; i++) mat[i][0] = 0; // Initializing first row with all 1s. // An empty string is subsequence of all. for (int j = 0; j <= n; j++) mat[0][j] = 1; // Fill mat[][] in bottom up manner for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { // If last characters don't match, // then value is same as the value // without last character in S. if (T.charAt(i - 1) != S.charAt(j - 1)) mat[i][j] = mat[i][j - 1]; // Else value is obtained considering two cases. // a) All substrings without last character in S // b) All substrings without last characters in // both. else mat[i][j] = mat[i][j - 1] + mat[i - 1][j - 1]; } } /* uncomment this to print matrix mat for (int i = 1; i <= m; i++, cout << endl) for (int j = 1; j <= n; j++) System.out.println ( mat[i][j] +" "); */ return mat[m][n] ; } // Driver code to check above method public static void main (String[] args) { String T = "ge"; String S = "geeksforgeeks"; System.out.println ( findSubsequenceCount(S, T)); } } // This code is contributed by vt_m

Output:

6

Time Complexity : O(m*n)

Auxiliary Space : O(m*n)

Since mat[i][j] accesses elements of current row and previous row only, we can optimize auxiliary space just by using two rows only reducing space from m*n to 2*n.

This article is contributed by **Utkarsh Trivedi**. 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.

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