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 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 Program int main() { char S[] = "digger"; char T[] = "biggerdiagram"; cout << maxPrefix(S, T) << endl; return 0; }

## 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 == t.length()) break; // If character match, increment // counter. if (t.charAt(i) == s.charAt(count)) count++; } return count; } // Driven Program 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

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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