Dadas dos strings s y t . La tarea es encontrar la longitud máxima de algún prefijo de la string S que aparece en la string t como subsecuencia.
Ejemplos:
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;
}
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
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.
?>
Javascript
<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>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA