Given two strings **s** and **t**. The task is to find maximum length of some prefix of the string S which occur in string t as subsequence.

**Examples :**

Input : s = "digger" t = "biggerdiagram" Output : 3digger biggerdiagram Prefix "dig" of s is longest subsequence in t. Input : s = "geeksforgeeks" t = "agbcedfeitk" Output : 4

A **simple solutions** is to consider all prefixes on by one and check if current prefix of s[] is a subsequence of t[] or not. Finally return length of the largest prefix.

An **efficient solution** is based on the fact that to find a prefix of length n, we must first find the prefix of length n – 1 and then look for s[n-1] in t. Similarly, to find a prefix of length n – 1, we must first find the prefix of length n – 2 and then look for s[n – 2] and so on.

Thus, we keep a counter which stores the current length of prefix found. We initialize it with 0 and begin with the first letter of s and keep iterating over t to find the occurrence of the first letter. As soon as we encounter the first letter of s we update the counter and look for second letter. We keep updating the counter and looking for next letter, until either the string s is found or there are no more letters in t.

Below is the implementation of this approach:

## C++

`// C++ program to find maximum ` `// length prefix of one string ` `// occur as subsequence in another ` `// string. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Return the maximum length ` `// prefix which is subsequence. ` `int` `maxPrefix(` `char` `s[], ` `char` `t[]) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Iterating string T. ` ` ` `for` `(` `int` `i = 0; i < ` `strlen` `(t); i++) ` ` ` `{ ` ` ` `// If end of string S. ` ` ` `if` `(count == ` `strlen` `(s)) ` ` ` `break` `; ` ` ` ` ` `// If character match, ` ` ` `// increment counter. ` ` ` `if` `(t[i] == s[count]) ` ` ` `count++; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driven Code ` `int` `main() ` `{ ` ` ` `char` `S[] = ` `"digger"` `; ` ` ` `char` `T[] = ` `"biggerdiagram"` `; ` ` ` ` ` `cout << maxPrefix(S, T) ` ` ` `<< endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find maximum ` `// length prefix of one string ` `// occur as subsequence in another ` `// string. ` `public` `class` `GFG { ` ` ` ` ` `// Return the maximum length ` ` ` `// prefix which is subsequence. ` ` ` `static` `int` `maxPrefix(String s, ` ` ` `String t) ` ` ` `{ ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// Iterating string T. ` ` ` `for` `(` `int` `i = ` `0` `; i < t.length(); i++) ` ` ` `{ ` ` ` `// If end of string S. ` ` ` `if` `(count == s.length()) ` ` ` `break` `; ` ` ` ` ` `// If character match, ` ` ` `// increment counter. ` ` ` `if` `(t.charAt(i) == s.charAt(count)) ` ` ` `count++; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `String S = ` `"digger"` `; ` ` ` `String T = ` `"biggerdiagram"` `; ` ` ` ` ` `System.out.println(maxPrefix(S, T)); ` ` ` `} ` `} ` `// This code is contributed by Sumit Ghosh ` |

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## Python 3

`# Python 3 program to find maximum ` `# length prefix of one string occur ` `# as subsequence in another string. ` ` ` ` ` `# Return the maximum length ` `# prefix which is subsequence. ` `def` `maxPrefix(s, t) : ` ` ` `count ` `=` `0` ` ` ` ` `# Iterating string T. ` ` ` `for` `i ` `in` `range` `(` `0` `,` `len` `(t)) : ` ` ` ` ` `# If end of string S. ` ` ` `if` `(count ` `=` `=` `len` `(s)) : ` ` ` `break` ` ` ` ` `# If character match, ` ` ` `# increment counter. ` ` ` `if` `(t[i] ` `=` `=` `s[count]) : ` ` ` `count ` `=` `count ` `+` `1` ` ` ` ` ` ` `return` `count ` ` ` ` ` `# Driver Code ` `S ` `=` `"digger"` `T ` `=` `"biggerdiagram"` ` ` `print` `(maxPrefix(S, T)) ` ` ` ` ` `# This code is contributed ` `# by Nikita Tiwari. ` |

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## C#

`// C# program to find maximum ` `// length prefix of one string ` `// occur as subsequence in ` `// another string. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Return the maximum length prefix ` ` ` `// which is subsequence. ` ` ` `static` `int` `maxPrefix(String s, ` ` ` `String t) ` ` ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// Iterating string T. ` ` ` `for` `(` `int` `i = 0; i < t.Length; i++) ` ` ` `{ ` ` ` `// If end of string S. ` ` ` `if` `(count == s.Length) ` ` ` `break` `; ` ` ` ` ` `// If character match, ` ` ` `// increment counter. ` ` ` `if` `(t[i] == s[count]) ` ` ` `count++; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `String S = ` `"digger"` `; ` ` ` `String T = ` `"biggerdiagram"` `; ` ` ` ` ` `Console.Write(maxPrefix(S, T)); ` ` ` `} ` `} ` ` ` `// This code is contributed by nitin mittal ` |

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## PHP

`<?php ` `// PHP program to find maximum ` `// length prefix of one string ` `// occur as subsequence in another ` `// string. ` ` ` `// Return the maximum length ` `// prefix which is subsequence. ` `function` `maxPrefix(` `$s` `, ` `$t` `) ` `{ ` ` ` `$count` `= 0; ` ` ` ` ` `// Iterating string T. ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `strlen` `(` `$t` `); ` `$i` `++) ` ` ` `{ ` ` ` `// If end of string S. ` ` ` `if` `(` `$count` `== ` `strlen` `(` `$s` `)) ` ` ` `break` `; ` ` ` ` ` `// If character match, ` ` ` `// increment counter. ` ` ` `if` `(` `$t` `[` `$i` `] == ` `$s` `[` `$count` `]) ` ` ` `$count` `++; ` ` ` `} ` ` ` ` ` `return` `$count` `; ` `} ` ` ` `// Driver Code ` `{ ` ` ` `$S` `= ` `"digger"` `; ` ` ` `$T` `= ` `"biggerdiagram"` `; ` ` ` ` ` `echo` `maxPrefix(` `$S` `, ` `$T` `) ; ` ` ` ` ` `return` `0; ` `} ` ` ` `// This code is contributed by nitin mittal. ` `?> ` |

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**Output :**

3

This article is contributed by **Anuj Chauhan**. 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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