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

**Examples:**

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

**Approach:** Create a recursive function such that it returns count of subsequences of *S* that match *T*. Here m is the length of T and n is length of S. This problem can be recursively defined as below.

- Given the string
*T*is an empty string, returning 1 as an empty string can be the subsequence of all. - Given the string
*S*is an empty string, returning 0 as no string can be the subsequence of an empty string. - If the last character of
*S*and*T*do not match, then remove the last character of*S*and call the recursive function again. Because the last character of S cannot be a part of the subsequence or remove it and check for other characters. - If the last character of
*S*match then there can be two possibilities, first there can be a subsequence where the last character of*S*is a part of it and second where it is not a part of the subsequence. So the required value will be the sum of both. Call the recursive function once with last character of both the strings removed and again with only last character of*S*removed.

**Base Cases:**

**Recursive cases:**

**Blue round rectangles represent accepted states or there are a subsequence and red round rectanges represent No subsequence can be formed.**

Since there are overlapping subproblems in the above recurrence result, Dynamic Programming approach can be applied to solve the above problem. Store the subproblems in a Hashmap or an array and return the value when the function is called again.

**Algorithm:**

- 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. - Initialize the first column with all 0s. An empty string can’t have another string as suhsequence
- Initialize the first row with all 1s. An empty string is subsequence of all.
- Fill the matrix in bottom up manner, i.e. all the sub problems of the current string is calculated first.
- Traverse the string
*T*from start to end. (counter is*i*) - For every iteration of the outer loop, Traverse the string
*S*from start to end. (counter is*j*) - If the character at
*i*th index of string*T*matches with*j*th character of string*S*, the value is obtained considering two cases. First, is all the substrings without last character in S and second is the substrings without last characters in both, i.e*mat[i+1][j] + mat[i][j]*. - Else the value will be same even if
*j*th character of*S*is removed, i.e. mat[i+1][j] - Print the value of
*mat[m-1][n-1]*as the answer.

## 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; ` `} ` |

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## 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 ` |

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## Python3

`# Python3 program to count number of times ` `# S appears as a subsequence in T ` `def` `findSubsequenceCount(S, T): ` ` ` ` ` `m ` `=` `len` `(T) ` ` ` `n ` `=` `len` `(S) ` ` ` ` ` `# 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). ` ` ` `mat ` `=` `[[` `0` `for` `_ ` `in` `range` `(n ` `+` `1` `)] ` ` ` `for` `__ ` `in` `range` `(m ` `+` `1` `)] ` ` ` ` ` `# Initializing first column with all 0s. x ` ` ` `# An empty string can't have another ` ` ` `# string as suhsequence ` ` ` `for` `i ` `in` `range` `(` `1` `, m ` `+` `1` `): ` ` ` `mat[i][` `0` `] ` `=` `0` ` ` ` ` `# Initializing first row with all 1s. ` ` ` `# An empty string is subsequence of all. ` ` ` `for` `j ` `in` `range` `(n ` `+` `1` `): ` ` ` `mat[` `0` `][j] ` `=` `1` ` ` ` ` `# Fill mat[][] in bottom up manner ` ` ` `for` `i ` `in` `range` `(` `1` `, m ` `+` `1` `): ` ` ` `for` `j ` `in` `range` `(` `1` `, n ` `+` `1` `): ` ` ` ` ` `# 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` `]) ` ` ` ` ` `return` `mat[m][n] ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `T ` `=` `"ge"` ` ` `S ` `=` `"geeksforgeeks"` ` ` `print` `(findSubsequenceCount(S, T)) ` ` ` `# This code is contributed ` `# by vibhu4agarwal ` |

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## C#

`// C# program to count number of times ` `// S appears as a subsequence in T ` `using` `System; ` ` ` `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[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++) ` ` ` `System.out.println ( mat[i][j] +" "); */` ` ` `return` `mat[m, n]; ` ` ` `} ` ` ` ` ` `// Driver code to check above method ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `string` `T = ` `"ge"` `; ` ` ` `string` `S = ` `"geeksforgeeks"` `; ` ` ` `Console.WriteLine(findSubsequenceCount(S, T)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m ` |

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## PHP

`<?php ` `// PHP program to count number of times ` `// S appears as a subsequence in T */ ` ` ` `function` `findSubsequenceCount(` `$S` `, ` `$T` `) ` `{ ` ` ` `$m` `= ` `strlen` `(` `$T` `); ` `$n` `= ` `strlen` `(` `$S` `); ` ` ` ` ` `// 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). ` ` ` `$mat` `= ` `array` `(` `array` `()); ` ` ` ` ` `// Initializing first column with all 0s. ` ` ` `// An empty string can't have another ` ` ` `// string as suhsequence ` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$m` `; ` `$i` `++) ` ` ` `$mat` `[` `$i` `][0] = 0; ` ` ` ` ` `// Initializing first row with all 1s. ` ` ` `// An empty string is subsequence of all. ` ` ` `for` `(` `$j` `= 0; ` `$j` `<= ` `$n` `; ` `$j` `++) ` ` ` `$mat` `[0][` `$j` `] = 1; ` ` ` ` ` `// Fill mat[][] in bottom up manner ` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$m` `; ` `$i` `++) ` ` ` `{ ` ` ` `for` `(` `$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` `][` `$n` `]; ` `} ` ` ` `// Driver Code ` `$T` `= ` `"ge"` `; ` `$S` `= ` `"geeksforgeeks"` `; ` `echo` `findSubsequenceCount(` `$S` `, ` `$T` `) . ` `"\n"` `; ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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**Output:**

6

**Complexity Analysis:**

**Time Complexity:**O(m*n).

Only one traversal of the matrix is needed, so the time Compelxiy is O(m*n)**Auxiliary Space:**O(m*n).

A matrix of size m*n is needed so the space complexity is O(m*n).

*Note:*Since mat[i][j] accesses elements of the 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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