Substring común más larga | DP-29

Dadas dos strings ‘X’ e ‘Y’, encuentre la longitud de la substring común más larga. 

Ejemplos: 

Entrada: X = «GeeksforGeeks», y = «GeeksQuiz» 
Salida : 5 
Explicación:
La substring común más larga es «Geeks» y tiene una longitud de 5.

Entrada: X = “abcdxyz”, y = “xyzabcd” 
Salida: 4 
Explicación:
La substring común más larga es “abcd” y tiene una longitud de 4.

Entrada: X = “zxabcdezy”, y = “yzabcdezx” 
Salida: 6 
Explicación:
La substring común más larga es “abcdez” y tiene una longitud de 6.

Enfoque:
Sean m y n las longitudes de la primera y segunda string respectivamente.

Una solución simple es considerar una por una todas las substrings de la primera string y para cada substring verificar si es una substring en la segunda string. Mantenga un registro de la substring de longitud máxima. Habrá O(m^2) substrings y podemos encontrar si una string es una substring en otra string en tiempo O(n) (Ver esto ). Entonces, la complejidad temporal general de este método sería O(n * m ² )

La programación dinámica se puede utilizar para encontrar la substring común más larga en tiempo O(m*n). La idea es encontrar la longitud del sufijo común más largo para todas las substrings de ambas strings y almacenar estas longitudes en una tabla. 

El sufijo común más largo tiene la siguiente propiedad de subestructura óptima. 
Si los últimos caracteres coinciden, entonces reducimos ambas longitudes en 1 
LCSuff(X, Y, m, n) = LCSuff(X, Y, m-1, n-1) + 1 si X[m-1] = Y[n -1] 
Si los últimos caracteres no coinciden, el resultado es 0, es decir, 
LCSuff(X, Y, m, n) = 0 if (X[m-1] != Y[n-1])
Ahora consideramos los sufijos de diferentes substrings que terminan en diferentes índices. 
El sufijo común más largo de longitud máxima es la substring común más larga. 
LCSubStr(X, Y, m, n) = Max(LCSuff(X, Y, i, j)) donde 1 <= i <= m y 1 <= j <= n 
 

A continuación se muestra la implementación iterativa de la solución anterior.  

/* Dynamic Programming solution to
   find length of the
   longest common substring */
#include <iostream>
#include <string.h>
using namespace std;
 
/* Returns length of longest
   common substring of X[0..m-1]
   and Y[0..n-1] */
int LCSubStr(char* X, char* Y, int m, int n)
{
    // Create a table to store
    // lengths of longest
    // common suffixes of substrings.  
    // Note that LCSuff[i][j] contains
    // length of longest common suffix
    // of X[0..i-1] and Y[0..j-1].
 
    int LCSuff[m + 1][n + 1];
    int result = 0; // To store length of the
                    // longest common substring
 
    /* Following steps build LCSuff[m+1][n+1] in
        bottom up fashion. */
    for (int i = 0; i <= m; i++)
    {
        for (int j = 0; j <= n; j++)
        {
            // The first row and first column
            // entries have no logical meaning,
            // they are used only for simplicity
            // of program
            if (i == 0 || j == 0)
                LCSuff[i][j] = 0;
 
            else if (X[i - 1] == Y[j - 1]) {
                LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1;
                result = max(result, LCSuff[i][j]);
            }
            else
                LCSuff[i][j] = 0;
        }
    }
    return result;
}
 
// Driver code
int main()
{
    char X[] = "OldSite:GeeksforGeeks.org";
    char Y[] = "NewSite:GeeksQuiz.com";
 
    int m = strlen(X);
    int n = strlen(Y);
 
    cout << "Length of Longest Common Substring is "
         << LCSubStr(X, Y, m, n);
    return 0;
}

//  Java implementation of
// finding length of longest
// Common substring using
// Dynamic Programming
class GFG {
    /*
       Returns length of longest common substring
       of X[0..m-1] and Y[0..n-1]
    */
    static int LCSubStr(char X[], char Y[],
                         int m, int n)
    {
        // Create a table to store
        // lengths of longest common
        // suffixes of substrings.
        // Note that LCSuff[i][j]
        // contains length of longest
        // common suffix of
        // X[0..i-1] and Y[0..j-1].
        // The first row and first
        // column entries have no
        // logical meaning, they are
        // used only for simplicity of program
        int LCStuff[][] = new int[m + 1][n + 1];
       
        // To store length of the longest
        // common substring
        int result = 0;
 
        // Following steps build
        // LCSuff[m+1][n+1] in bottom up fashion
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    LCStuff[i][j] = 0;
                else if (X[i - 1] == Y[j - 1])
                {
                    LCStuff[i][j]
                        = LCStuff[i - 1][j - 1] + 1;
                    result = Integer.max(result,
                                         LCStuff[i][j]);
                }
                else
                    LCStuff[i][j] = 0;
            }
        }
        return result;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String X = "OldSite:GeeksforGeeks.org";
        String Y = "NewSite:GeeksQuiz.com";
 
        int m = X.length();
        int n = Y.length();
 
        System.out.println(LCSubStr(X.toCharArray(),
                                    Y.toCharArray(), m,
                       n));
    }
}
 
// This code is contributed by Sumit Ghosh

# Python3 implementation of Finding
# Length of Longest Common Substring
 
# Returns length of longest common
# substring of X[0..m-1] and Y[0..n-1]
 
 
def LCSubStr(X, Y, m, n):
 
    # Create a table to store lengths of
    # longest common suffixes of substrings.
    # Note that LCSuff[i][j] contains the
    # length of longest common suffix of
    # X[0...i-1] and Y[0...j-1]. The first
    # row and first column entries have no
    # logical meaning, they are used only
    # for simplicity of the program.
 
    # LCSuff is the table with zero
    # value initially in each cell
    LCSuff = [[0 for k in range(n+1)] for l in range(m+1)]
 
    # To store the length of
    # longest common substring
    result = 0
 
    # Following steps to build
    # LCSuff[m+1][n+1] in bottom up fashion
    for i in range(m + 1):
        for j in range(n + 1):
            if (i == 0 or j == 0):
                LCSuff[i][j] = 0
            elif (X[i-1] == Y[j-1]):
                LCSuff[i][j] = LCSuff[i-1][j-1] + 1
                result = max(result, LCSuff[i][j])
            else:
                LCSuff[i][j] = 0
    return result
 
 
# Driver Code
X = 'OldSite:GeeksforGeeks.org'
Y = 'NewSite:GeeksQuiz.com'
 
m = len(X)
n = len(Y)
 
print('Length of Longest Common Substring is',
      LCSubStr(X, Y, m, n))
 
# This code is contributed by Soumen Ghosh

// C# implementation of finding length of longest
// Common substring using Dynamic Programming
using System;
 
class GFG {
 
    // Returns length of longest common
    // substring of X[0..m-1] and Y[0..n-1]
    static int LCSubStr(string X, string Y, int m, int n)
    {
 
        // Create a table to store lengths of
        // longest common suffixes of substrings.
        // Note that LCSuff[i][j] contains length
        // of longest common suffix of X[0..i-1]
        // and Y[0..j-1]. The first row and first
        // column entries have no logical meaning,
        // they are used only for simplicity of
        // program
        int[, ] LCStuff = new int[m + 1, n + 1];
 
        // To store length of the longest common
        // substring
        int result = 0;
 
        // Following steps build LCSuff[m+1][n+1]
        // in bottom up fashion
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    LCStuff[i, j] = 0;
                else if (X[i - 1] == Y[j - 1])
                {
                    LCStuff[i, j]
                        = LCStuff[i - 1, j - 1] + 1;
 
                    result
                        = Math.Max(result, LCStuff[i, j]);
                }
                else
                    LCStuff[i, j] = 0;
            }
        }
 
        return result;
    }
 
    // Driver Code
    public static void Main()
    {
        String X = "OldSite:GeeksforGeeks.org";
        String Y = "NewSite:GeeksQuiz.com";
 
        int m = X.Length;
        int n = Y.Length;
 
        Console.Write("Length of Longest Common"
                      + " Substring is "
                      + LCSubStr(X, Y, m, n));
    }
}
 
// This code is contributed by Sam007.

<?php
// Dynamic Programming solution to find
// length of the longest common substring
 
// Returns length of longest common
// substring of X[0..m-1] and Y[0..n-1]
function LCSubStr($X, $Y, $m, $n)
{
    // Create a table to store lengths of
    // longest common suffixes of substrings.
    // Notethat LCSuff[i][j] contains length
    // of longest common suffix of X[0..i-1]
    // and Y[0..j-1]. The first row and
    // first column entries have no logical
    // meaning, they are used only for
    // simplicity of program
    $LCSuff = array_fill(0, $m + 1,
              array_fill(0, $n + 1, NULL));
    $result = 0; // To store length of the
                 // longest common substring
 
    // Following steps build LCSuff[m+1][n+1]
    // in bottom up fashion.
    for ($i = 0; $i <= $m; $i++)
    {
        for ($j = 0; $j <= $n; $j++)
        {
            if ($i == 0 || $j == 0)
                $LCSuff[$i][$j] = 0;
 
            else if ($X[$i - 1] == $Y[$j - 1])
            {
                $LCSuff[$i][$j] = $LCSuff[$i - 1][$j - 1] + 1;
                $result = max($result,
                              $LCSuff[$i][$j]);
            }
            else $LCSuff[$i][$j] = 0;
        }
    }
    return $result;
}
 
// Driver Code
$X = "OldSite:GeeksforGeeks.org";
$Y = "NewSite:GeeksQuiz.com";
 
$m = strlen($X);
$n = strlen($Y);
 
echo "Length of Longest Common Substring is " .
                      LCSubStr($X, $Y, $m, $n);
                       
// This code is contributed by ita_c
?>

<script>
 
// JavaScript implementation of
// finding length of longest
// Common substring using
// Dynamic Programming
 
    /*
     Returns length of longest common
     substring of X[0..m-1] and Y[0..n-1]
     */
    function LCSubStr( X,  Y , m , n) {
        // Create a table to store
        // lengths of longest common
        // suffixes of substrings.
        // Note that LCSuff[i][j]
        // contains length of longest
        // common suffix of
        // X[0..i-1] and Y[0..j-1].
        // The first row and first
        // column entries have no
        // logical meaning, they are
        // used only for simplicity of program
         
        var LCStuff =
        Array(m + 1).fill().map(()=>Array(n + 1).fill(0));
 
        // To store length of the longest
        // common substring
        var result = 0;
 
        // Following steps build
        // LCSuff[m+1][n+1] in bottom up fashion
        for (i = 0; i <= m; i++) {
            for (j = 0; j <= n; j++) {
                if (i == 0 || j == 0)
                    LCStuff[i][j] = 0;
                else if (X[i - 1] == Y[j - 1]) {
                    LCStuff[i][j] = LCStuff[i - 1][j - 1] + 1;
                    result = Math.max(result, LCStuff[i][j]);
                } else
                    LCStuff[i][j] = 0;
            }
        }
        return result;
    }
 
    // Driver Code
     
        var X = "OldSite:GeeksforGeeks.org";
        var Y = "NewSite:GeeksQuiz.com";
 
        var m = X.length;
        var n = Y.length;
 
        document.write("Length of Longest Common Substring is " +
        LCSubStr(X, Y, m, n));
 
// This code contributed by Rajput-Ji
 
</script>

Length of Longest Common Substring is 10

Tiempo Complejidad: O(m*n) 
Espacio Auxiliar: O(m*n)

Otro enfoque: (enfoque de espacio optimizado).
En el enfoque anterior, solo usamos la última fila de la array 2D, por lo que podemos optimizar el espacio usando 
una array 2D de dimensión 2*(min(n,m)).

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the
// longest LCS
int LCSubStr(string s, string t, int n, int m)
{
   
    // Create DP table
    int dp[2][m + 1];
    int res = 0;
 
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (s[i - 1] == t[j - 1]) {
                dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;
                if (dp[i % 2][j] > res)
                    res = dp[i % 2][j];
            }
            else
                dp[i % 2][j] = 0;
        }
    }
    return res;
}
 
// Driver Code
int main()
{
    string X = "OldSite:GeeksforGeeks.org";
    string Y = "NewSite:GeeksQuiz.com";
 
    int m = X.length();
    int n = Y.length();
 
    cout << LCSubStr(X, Y, m, n);
    return 0;
    cout << "GFG!";
    return 0;
}
 
// This code is contributed by rajsanghavi9.

// Java implementation of the above approach
 
class GFG
{
   
    // Function to find the length of the
    // longest LCS
    static int LCSubStr(String s,String t,
                        int n,int m)
    { 
       
        // Create DP table
        int dp[][]=new int[2][m+1];
        int res=0;
      
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(s.charAt(i-1)==t.charAt(j-1))
                {
                    dp[i%2][j]=dp[(i-1)%2][j-1]+1;
                    if(dp[i%2][j]>res)
                        res=dp[i%2][j];
                }
                else dp[i%2][j]=0;
            }
        }
        return res;
    }
   
    // Driver Code
    public static void main (String[] args)
    {
        String X="OldSite:GeeksforGeeks.org";
        String Y="NewSite:GeeksQuiz.com";
         
        int m=X.length();
        int n=Y.length();
         
        // Function call
        System.out.println(LCSubStr(X,Y,m,n));
         
    }
}

# Python implementation of the above approach
 
# Function to find the length of the
# longest LCS
def LCSubStr(s, t, n, m):
   
    # Create DP table
    dp = [[0 for i in range(m + 1)] for j in range(2)]
    res = 0
     
    for i in range(1,n + 1):
        for j in range(1,m + 1):
            if(s[i - 1] == t[j - 1]):
                dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1
                if(dp[i % 2][j] > res):
                    res = dp[i % 2][j]
            else:
                dp[i % 2][j] = 0
    return res
 
# Driver Code
X = "OldSite:GeeksforGeeks.org"
Y = "NewSite:GeeksQuiz.com"
m = len(X)
n = len(Y)
 
# Function call
print(LCSubStr(X,Y,m,n))
 
# This code is contributed by avanitrachhadiya2155

// C# implementation of the above approach
using System;
public class GFG
{
 
  // Function to find the length of the
  // longest LCS
  static int LCSubStr(string s,string t,
                      int n,int m)
  { 
 
    // Create DP table
    int[,] dp = new int[2, m + 1];
    int res = 0;
 
    for(int i = 1; i <= n; i++)
    {
      for(int j = 1; j <= m; j++)
      {
        if(s[i - 1] == t[j - 1])
        {
          dp[i % 2, j] = dp[(i - 1) % 2, j - 1] + 1;
          if(dp[i % 2, j] > res)
            res = dp[i % 2, j];
        }
        else dp[i % 2, j] = 0;
      }
    }
    return res;
  }
 
  // Driver Code
  static public void Main (){
    string X = "OldSite:GeeksforGeeks.org";
    string Y = "NewSite:GeeksQuiz.com";
 
    int m = X.Length;
    int n = Y.Length;
 
    // Function call
    Console.WriteLine(LCSubStr(X,Y,m,n));
  }
}
 
// This code is contributed by rag2127

<script>
// JavaScript implementation of the above approach
 
    // Function to find the length of the
    // longest LCS
    function LCSubStr(s, t, n, m)
    { 
       
        // Create DP table
        var dp = Array(2).fill().map(()=>Array(m+ 1).fill(0));
        var res = 0;
      
        for(var i = 1; i <= n; i++)
        {
            for(var j = 1; j <= m; j++)
            {
                if(s.charAt(i - 1) == t.charAt(j - 1))
                {
                    dp[i % 2][j] = dp[(i - 1) % 2][j - 1] + 1;
                    if(dp[i % 2][j] > res)
                        res = dp[i % 2][j];
                }
                else dp[i % 2][j] = 0;
            }
        }
        return res;
    }
   
    // Driver Code
        var X = "OldSite:GeeksforGeeks.org";
        var Y = "NewSite:GeeksQuiz.com";
         
        var m = X.length;
        var n = Y.length;
         
        // Function call
        document.write(LCSubStr(X, Y, m, n));
 
// This code is contributed by shivanisinghss2110
</script>

10

Complejidad temporal: O(n*m)
Espacio auxiliar: O(min(m,n))

Otro enfoque: (usando la recursividad) 
Aquí está la solución recursiva del enfoque anterior. 

// C++ program using to find length of the
// longest common substring  recursion
#include <iostream>
 
using namespace std;
 
string X, Y;
 
// Returns length of function f
// or longest common substring
// of X[0..m-1] and Y[0..n-1]
int lcs(int i, int j, int count)
{
 
    if (i == 0 || j == 0)
        return count;
 
    if (X[i - 1] == Y[j - 1]) {
        count = lcs(i - 1, j - 1, count + 1);
    }
    count = max(count,
                max(lcs(i, j - 1, 0),
                    lcs(i - 1, j, 0)));
    return count;
}
 
// Driver code
int main()
{
    int n, m;
 
    X = "abcdxyz";
    Y = "xyzabcd";
 
    n = X.size();
    m = Y.size();
 
    cout << lcs(n, m, 0);
 
    return 0;
}

// Java program using to find length of the
// longest common substring recursion
 
class GFG {
 
    static String X, Y;
    // Returns length of function
    // for longest common
    // substring of X[0..m-1] and Y[0..n-1]
    static int lcs(int i, int j, int count)
    {
 
        if (i == 0 || j == 0)
        {
            return count;
        }
 
        if (X.charAt(i - 1)
            == Y.charAt(j - 1))
        {
            count = lcs(i - 1, j - 1, count + 1);
        }
        count = Math.max(count,
                         Math.max(lcs(i, j - 1, 0),
                                  lcs(i - 1, j, 0)));
        return count;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int n, m;
        X = "abcdxyz";
        Y = "xyzabcd";
 
        n = X.length();
        m = Y.length();
 
        System.out.println(lcs(n, m, 0));
    }
}
// This code is contributed by Rajput-JI

# Python3 program using to find length of
# the longest common substring recursion
 
# Returns length of function for longest
# common substring of X[0..m-1] and Y[0..n-1]
 
 
def lcs(i, j, count):
 
    if (i == 0 or j == 0):
        return count
 
    if (X[i - 1] == Y[j - 1]):
        count = lcs(i - 1, j - 1, count + 1)
 
    count = max(count, max(lcs(i, j - 1, 0),
                           lcs(i - 1, j, 0)))
 
    return count
 
 
# Driver code
if __name__ == "__main__":
 
    X = "abcdxyz"
    Y = "xyzabcd"
 
    n = len(X)
    m = len(Y)
 
    print(lcs(n, m, 0))
 
# This code is contributed by Ryuga

// C# program using to find length
// of the longest common substring
// recursion
using System;
 
class GFG {
    static String X, Y;
 
    // Returns length of function for
    // longest common substring of
    // X[0..m-1] and Y[0..n-1]
    static int lcs(int i, int j, int count)
    {
 
        if (i == 0 || j == 0) {
            return count;
        }
 
        if (X[i - 1] == Y[j - 1]) {
            count = lcs(i - 1, j - 1, count + 1);
        }
        count = Math.Max(count, Math.Max(lcs(i, j - 1, 0),
                                         lcs(i - 1, j, 0)));
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int n, m;
        X = "abcdxyz";
        Y = "xyzabcd";
 
        n = X.Length;
        m = Y.Length;
 
        Console.Write(lcs(n, m, 0));
    }
}
 
// This code is contributed by Rajput-JI

<?php
// PHP program using to find length of the
// longest common substring recursion
 
// Returns length of function for
// longest common substring of
// X[0..m-1] and Y[0..n-1]
function lcs($i, $j, $count, &$X, &$Y)
{
    if ($i == 0 || $j == 0)
        return $count;
         
    if ($X[$i - 1] == $Y[$j - 1])
    {
        $count = lcs($i - 1, $j - 1,
                     $count + 1, $X, $Y);
    }
        $count = max($count, lcs($i, $j - 1, 0, $X, $Y),
                             lcs($i - 1, $j, 0, $X, $Y));
    return $count;
}
 
// Driver code
$X = "abcdxyz";
$Y = "xyzabcd";
 
$n = strlen($X);
$m = strlen($Y);
 
echo lcs($n, $m, 0, $X, $Y);
 
// This code is contributed
// by rathbhupendra
?>

<script>
    // Javascript program using to find length of the
    // longest common substring  recursion
    let X, Y;
  
    // Returns length of function f
    // or longest common substring
    // of X[0..m-1] and Y[0..n-1]
    function lcs(i, j, count)
    {
      
        if (i == 0 || j == 0)
            return count;
      
        if (X[i - 1] == Y[j - 1]) {
            count = lcs(i - 1, j - 1, count + 1);
        }
        count = Math.max(count,
                    Math.max(lcs(i, j - 1, 0),
                        lcs(i - 1, j, 0)));
        return count;
    }
     
    let n, m;
  
    X = "abcdxyz";
    Y = "xyzabcd";
  
    n = X.length;
    m = Y.length;
  
    document.write(lcs(n, m, 0));
     
    // This code is contributed by divyeshrabadiya07.
</script>

4

Optimización máxima del espacio :

1. En este método, usaremos la recursividad para encontrar el prefijo más largo de todas las substrings posibles.
2. Let   da la longitud del sufijo común más largo a partir de los índices i , j de strings X , Y respectivamente.
3. Entonces la función se puede definir como:
   
4. En esta recursión podemos ver que la función tiene solo una dependencia, lo que significa que podemos memorizar solo el cálculo anterior si hacemos nuestro cálculo en un orden específico.
5. Considere la siguiente tabla donde memorizamos las soluciones:

Necesitamos encontrar la solución en diagonal hacia arriba. En este ejemplo en particular:

• primera diagonal
  • (4, 0)
• segunda diagonal
  • (4, 1)
  • (3, 0)
• tercera diagonal
  • (4, 2)
  • (3, 1)
  • (2, 0)
• …

De esta manera, necesitamos recordar solo el cálculo anterior.

# Python code for the above approach
from functools import lru_cache
from operator import itemgetter
 
def longest_common_substring(x: str, y: str) -> (int, int, int):
     
    # function to find the longest common substring
 
    # Memorizing with maximum size of the memory as 1
    @lru_cache(maxsize=1) 
     
    # function to find the longest common prefix
    def longest_common_prefix(i: int, j: int) -> int:
       
        if 0 <= i < len(x) and 0 <= j < len(y) and x[i] == y[j]:
            return 1 + longest_common_prefix(i + 1, j + 1)
        else:
            return 0
 
    # diagonally computing the subproblems
    # to decrease memory dependency
    def digonal_computation():
         
        # upper right triangle of the 2D array
        for k in range(len(x)):       
            yield from ((longest_common_prefix(i, j), i, j)
                        for i, j in zip(range(k, -1, -1),
                                    range(len(y) - 1, -1, -1)))
         
        # lower left triangle of the 2D array
        for k in range(len(y)):       
            yield from ((longest_common_prefix(i, j), i, j)
                        for i, j in zip(range(k, -1, -1),
                                    range(len(x) - 1, -1, -1)))
 
    # returning the maximum of all the subproblems
    return max(digonal_computation(), key=itemgetter(0), default=(0, 0, 0))
 
# Driver Code
if __name__ == '__main__':
    x: str = 'GeeksforGeeks'
    y: str = 'GeeksQuiz'
    length, i, j = longest_common_substring(x, y)
    print(f'length: {length}, i: {i}, j: {j}')
    print(f'x substring: {x[i: i + length]}')
    print(f'y substring: {y[j: j + length]}')

length: 5, i: 0, j: 0
x substring: Geeks
y substring: Geeks

Complejidad de tiempo : 
Complejidad de espacio : 

Ejercicio: La solución anterior imprime solo la longitud de la substring común más larga. Extienda la solución para imprimir también la substring.