# Find number of times a string occurs as a subsequence in given string

Given two strings, find the number of times the second string occurs in the first string, whether continuous or discontinuous.

Examples:

```Input:
string a = "GeeksforGeeks"
string b = "Gks"

Output: 4

Explanation:
The four strings are - (Check characters marked in bold)
GeeksforGeeks
GeeksforGeeks
GeeksforGeeks
GeeksforGeeks
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

If we carefully analyze the given problem, we can see that it can be easily divided into sub-problems. The idea is to process all characters of both strings one by one staring from either from left or right side. Let us traverse from right corner, there are two possibilities for every pair of character being traversed.

```m: Length of str1 (first string)
n: Length of str2 (second string)

If last characters of two strings are same,
1. We consider last characters and get count for remaining
strings. So we recur for lengths m-1 and n-1.
2. We can ignore last character of first string and
recurse for lengths m-1 and n.
else
If last characters are not same,
We ignore last character of first string and
recurse for lengths m-1 and n.
```

Below is C++ implementation of above Naive recursive solution –

```// A Naive recursive C++ program to find the number of
// times the second string occurs in the first string,
// whether continuous or discontinuous
#include <iostream>
using namespace std;

// Recursive function to find the number of times
// the second string occurs in the first string,
// whether continuous or discontinuous
int count(string a, string b, int m, int n)
{
// If both first and second string is empty,
// or if second string is empty, return 1
if ((m == 0 && n == 0) || n == 0)
return 1;

// If only first string is empty and second
// string is not empty, return 0
if (m == 0)
return 0;

// If last characters are same
// Recur for remaining strings by
// 1. considering last characters of both strings
// 2. ignoring last character of first string
if (a[m - 1] == b[n - 1])
return count(a, b, m - 1, n - 1) +
count(a, b, m - 1, n);
else
// If last characters are different, ignore
// last char of first string and recur for
// remaining string
return count(a, b, m - 1, n);
}

// Driver code
int main()
{
string a = "GeeksforGeeks";
string b = "Gks";

cout << count(a, b, a.size(), b.size()) << endl;

return 0;
}
```

Output:

```4
```

The time complexity of above solution is exponential. If we carefully analyze, we can see that many sub-problems are solved again and again. Since same sub-problems are called again, this problem has Overlapping sub-problems property. So the problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming problems, re-computations of same sub-problems can be avoided by constructing a temporary array that stores results of sub-problems.

Below is its C++ implementation using Dynamic Programming –

```// A Dynamic Programming based C++ program to find the
// number of times the second string occurs in the first
// string, whether continuous or discontinuous
#include <iostream>
using namespace std;

// Iterative DP function to find the number of times
// the second string occurs in the first string,
// whether continuous or discontinuous
int count(string a, string b)
{
int m = a.length();
int n = b.length();

// Create a table to store results of sub-problems
int lookup[m + 1][n + 1] = { { 0 } };

// If first string is empty
for (int i = 0; i <= n; ++i)
lookup[0][i] = 0;

// If second string is empty
for (int i = 0; i <= m; ++i)
lookup[i][0] = 1;

// Fill lookup[][] in bottom up manner
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
// If last characters are same, we have two
// options -
// 1. consider last characters of both strings
//    in solution
// 2. ignore last character of first string
if (a[i - 1] == b[j - 1])
lookup[i][j] = lookup[i - 1][j - 1] +
lookup[i - 1][j];

else
// If last character are different, ignore
// last character of first string
lookup[i][j] = lookup[i - 1][j];
}
}

return lookup[m][n];
}

// Driver code
int main()
{
string a = "GeeksforGeeks";
string b = "Gks";

cout << count(a, b);

return 0;
}
```

Output:

```4
```

Time complexity of above solutions is O(MN).
Auxiliary space used by the program is O(MN).

This article is contributed by Aditya Goel. 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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