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

A simple solutions is to consider all prefixes one 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++ 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;
}

// 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

# 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# 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

<?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.
?>

<script>

// JavaScript 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)
    {
        let count = 0;
    
        // Iterating string T.
        for (let 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

    let S = "digger";
    let T = "biggerdiagram";
    
    document.write(maxPrefix(S, T)); 
        
</script>

3

Time complexity: O(n)

Space complexity: O(1)