Maximum length prefix of one string that occurs as subsequence in another

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 : 3
digger
biggerdiagram
Prefix "dig" of s is longest subsequence in t.

Input : s = "geeksforgeeks"
t = "agbcedfeitk"
Output : 4

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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; }

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;     }            // 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

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.

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 == t.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

PHP



Output :

3

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Improved By : nitin mittal, nidhi_biet