一个字符串的最大长度前缀，在另一个字符串中作为子序列出现

给定两个字符串s和t 。任务是找到在字符串t中作为子序列出现的字符串S的某些前缀的最大长度。

例子 ：

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

一个简单的解决方案是考虑所有前缀加一个，然后检查s []的当前前缀是否为t []的子序列。最后返回最大前缀的长度。

一个有效的解决方案基于以下事实：要找到长度为n的前缀，我们必须首先找到长度为n – 1的前缀，然后在t中寻找s [n-1]。同样，要找到长度为n – 1的前缀，我们必须首先找到长度为n – 2的前缀，然后寻找s [n – 2]，依此类推。
因此，我们保留一个计数器，该计数器存储找到的前缀的当前长度。我们将其初始化为0，然后从s的第一个字母开始，并在t上进行迭代，以查找第一个字母的出现。一旦遇到s的第一个字母，我们就会更新计数器并寻找第二个字母。我们会不断更新计数器并寻找下一个字母，直到找到字符串s或t中没有更多字母为止。

以下是此方法的实现：

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 == 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

# 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 == 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

输出 ：

3