Dadas 3 strings de todas con una longitud < 100, la tarea es encontrar la subsecuencia común más larga en las tres secuencias dadas.
Ejemplos:
Input : str1 = "geeks"
str2 = "geeksfor"
str3 = "geeksforgeeks"
Output : 5
Longest common subsequence is "geeks"
i.e., length = 5
Input : str1 = "abcd1e2"
str2 = "bc12ea"
str3 = "bd1ea"
Output : 3
Longest common subsequence is "b1e"
i.e. length = 3.
Este problema es simplemente una extensión de LCS
Sean las secuencias de entrada X[0..m-1], Y[0..n-1] y Z[0..o-1] de longitudes m, n y o respectivamente . Y sean L(X[0..m-1], Y[0..n-1], Z[0..o-1]) las longitudes de LCS de las tres secuencias X, Y y Z.
A continuación se muestra la implementación:
The idea is to take a 3D array to store the
length of common subsequence in all 3 given
sequences i. e., L[m + 1][n + 1][o + 1]
1- If any of the string is empty then there
is no common subsequence at all then
L[i][j][k] = 0
2- If the characters of all sequences match
(or X[i] == Y[j] ==Z[k]) then
L[i][j][k] = 1 + L[i-1][j-1][k-1]
3- If the characters of both sequences do
not match (or X[i] != Y[j] || X[i] != Z[k]
|| Y[j] !=Z[k]) then
L[i][j][k] = max(L[i-1][j][k],
L[i][j-1][k],
L[i][j][k-1])
A continuación se muestra la implementación de la idea anterior.
C++
// C++ program to find LCS of three strings
#include<bits/stdc++.h>
using namespace std;
/* Returns length of LCS for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
int lcsOf3( string X, string Y, string Z, int m,
int n, int o)
{
int L[m+1][n+1][o+1];
/* Following steps build L[m+1][n+1][o+1] in
bottom up fashion. Note that L[i][j][k]
contains length of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1]*/
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
for (int k=0; k<=o; k++)
{
if (i == 0 || j == 0||k==0)
L[i][j][k] = 0;
else if (X[i-1] == Y[j-1] && X[i-1]==Z[k-1])
L[i][j][k] = L[i-1][j-1][k-1] + 1;
else
L[i][j][k] = max(max(L[i-1][j][k],
L[i][j-1][k]),
L[i][j][k-1]);
}
}
}
/* L[m][n][o] contains length of LCS for
X[0..n-1] and Y[0..m-1] and Z[0..o-1]*/
return L[m][n][o];
}
/* Driver program to test above function */
int main()
{
string X = "AGGT12";
string Y = "12TXAYB";
string Z = "12XBA";
int m = X.length();
int n = Y.length();
int o = Z.length();
cout << "Length of LCS is " << lcsOf3(X, Y,
Z, m, n, o);
return 0;
}
Java
// Java program to find LCS of three strings
public class LCS_3Strings {
/* Returns length of LCS for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
static int lcsOf3(String X, String Y, String Z, int m,
int n, int o)
{
int[][][] L = new int[m+1][n+1][o+1];
/* Following steps build L[m+1][n+1][o+1] in
bottom up fashion. Note that L[i][j][k]
contains length of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1]*/
for (int i=0; i<=m; i++)
{
for (int j=0; j<=n; j++)
{
for (int k=0; k<=o; k++)
{
if (i == 0 || j == 0||k==0)
L[i][j][k] = 0;
else if (X.charAt(i - 1) == Y.charAt(j - 1)
&& X.charAt(i - 1)==Z.charAt(k - 1))
L[i][j][k] = L[i-1][j-1][k-1] + 1;
else
L[i][j][k] = Math.max(Math.max(L[i-1][j][k],
L[i][j-1][k]),
L[i][j][k-1]);
}
}
}
/* L[m][n][o] contains length of LCS for
X[0..n-1] and Y[0..m-1] and Z[0..o-1]*/
return L[m][n][o];
}
/* Driver program to test above function */
public static void main(String args[])
{
String X = "AGGT12";
String Y = "12TXAYB";
String Z = "12XBA";
int m = X.length();
int n = Y.length();
int o = Z.length();
System.out.println("Length of LCS is " +
lcsOf3(X, Y,Z, m, n, o));
}
}
// This code is contributed by Sumit Ghosh
Python3
# Python program to find
# LCS of three strings
# Returns length of LCS
# for X[0..m-1], Y[0..n-1]
# and Z[0..o-1]
def lcsOf3(X, Y, Z, m, n, o):
L = [[[0 for i in range(o+1)] for j in range(n+1)]
for k in range(m+1)]
''' Following steps build L[m+1][n+1][o+1] in
bottom up fashion. Note that L[i][j][k]
contains length of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1] '''
for i in range(m+1):
for j in range(n+1):
for k in range(o+1):
if (i == 0 or j == 0 or k == 0):
L[i][j][k] = 0
elif (X[i-1] == Y[j-1] and
X[i-1] == Z[k-1]):
L[i][j][k] = L[i-1][j-1][k-1] + 1
else:
L[i][j][k] = max(max(L[i-1][j][k],
L[i][j-1][k]),
L[i][j][k-1])
# L[m][n][o] contains length of LCS for
# X[0..n-1] and Y[0..m-1] and Z[0..o-1]
return L[m][n][o]
# Driver program to test above function
X = 'AGGT12'
Y = '12TXAYB'
Z = '12XBA'
m = len(X)
n = len(Y)
o = len(Z)
print('Length of LCS is', lcsOf3(X, Y, Z, m, n, o))
# This code is contributed by Soumen Ghosh.
C#
// C# program to find
// LCS of three strings
using System;
class GFG
{
/* Returns length of LCS
for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
static int lcsOf3(String X, String Y,
String Z, int m,
int n, int o)
{
int [,,]L = new int[m + 1,
n + 1, o + 1];
/* Following steps build
L[m+1][n+1][o+1] in bottom
up fashion. Note that
L[i][j][k] contains length
of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1]*/
for (int i = 0; i <= m; i++)
{
for (int j = 0; j <= n; j++)
{
for (int k = 0; k <= o; k++)
{
if (i == 0 ||
j == 0 || k == 0)
L[i, j, k] = 0;
else if (X[i - 1] == Y[j - 1] &&
X[i - 1] == Z[k - 1])
L[i, j, k] = L[i - 1,
j - 1,
k - 1] + 1;
else
L[i, j, k] = Math.Max(Math.Max(L[i - 1, j, k],
L[i, j - 1, k]),
L[i, j, k - 1]);
}
}
}
/* L[m][n][o] contains length
of LCS for X[0..n-1] and
Y[0..m-1] and Z[0..o-1]*/
return L[m, n, o];
}
// Driver Code
public static void Main()
{
string X = "AGGT12";
string Y = "12TXAYB";
string Z = "12XBA";
int m = X.Length;
int n = Y.Length;
int o = Z.Length;
Console.Write("Length of LCS is " +
lcsOf3(X, Y, Z, m, n, o));
}
}
// This code is contributed
// by shiv_bhakt.
PHP
<?php
// PHP program to find
// LCS of three strings
/* Returns length of LCS for
X[0..m-1], Y[0..n-1] and Z[0..o-1] */
function lcsOf3($X, $Y, $Z,
$m, $n, $o)
{
$L[$m + 1][$n + 1][$o + 1] = array(array(array()));
/* Following steps build
L[m+1][n+1][o+1] in bottom
up fashion. Note that
L[i][j][k] contains length
of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1]*/
for ($i = 0; $i <= $m; $i++)
{
for ($j = 0; $j <= $n; $j++)
{
for ($k = 0; $k <= $o; $k++)
{
if ($i == 0 || $j == 0||$k == 0)
$L[$i][$j][$k] = 0;
else if ($X[$i - 1] == $Y[$j - 1] &&
$X[$i - 1] == $Z[$k - 1])
$L[$i][$j][$k] = $L[$i - 1][$j - 1][$k - 1] + 1;
else
$L[$i][$j][$k] = max(max($L[$i - 1][$j][$k],
$L[$i][$j - 1][$k]),
$L[$i][$j][$k - 1]);
}
}
}
/* L[m][n][o] contains length
of LCS for X[0..n-1] and
Y[0..m-1] and Z[0..o-1]*/
return $L[$m][$n][$o];
}
// Driver code
$X = "AGGT12";
$Y = "12TXAYB";
$Z = "12XBA";
$m = strlen($X);
$n = strlen($Y);
$o = strlen($Z);
echo "Length of LCS is " .
lcsOf3($X, $Y, $Z,
$m, $n, $o);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Javascript program to find LCS of three strings
/* Returns length of LCS for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
function lcsOf3(X,Y,Z,m,n,o)
{
let L = new Array(m + 1);
for(let i = 0; i < m + 1; i++)
{
L[i] = new Array(n + 1);
for(let j = 0; j < n + 1; j++)
{
L[i][j] = new Array(o + 1);
for(let k = 0; k < o + 1; k++)
{
L[i][j][k] = 0;
}
}
}
/* Following steps build L[m+1][n+1][o+1] in
bottom up fashion. Note that L[i][j][k]
contains length of LCS of X[0..i-1] and
Y[0..j-1] and Z[0.....k-1]*/
for (let i = 0; i <= m; i++)
{
for (let j = 0; j <= n; j++)
{
for (let k = 0; k <= o; k++)
{
if (i == 0 || j == 0 || k == 0)
L[i][j][k] = 0;
else if (X[i - 1] == Y[j - 1]
&& X[i - 1] == Z[k - 1])
L[i][j][k] = L[i - 1][j - 1][k - 1] + 1;
else
L[i][j][k] = Math.max(Math.max(L[i - 1][j][k],
L[i][j - 1][k]),
L[i][j][k - 1]);
}
}
}
/* L[m][n][o] contains length of LCS for
X[0..n-1] and Y[0..m-1] and Z[0..o-1]*/
return L[m][n][o];
}
/* Driver program to test above function */
let X = "AGGT12";
let Y = "12TXAYB";
let Z = "12XBA";
let m = X.length;
let n = Y.length;
let o = Z.length;
document.write("Length of LCS is " +
lcsOf3(X, Y,Z, m, n, o));
// This code is contributed by avanitrachhadiya2155
</script>
Producción
Length of LCS is 2
Otro enfoque: (usando la recursividad)
C++
// C++ program to find LCS of three strings
#include<bits/stdc++.h>
using namespace std;
string X = "AGGT12";
string Y = "12TXAYB";
string Z = "12XBA";
int dp[100][100][100];
/* Returns length of LCS for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
int lcsOf3(int i, int j,int k)
{
if(i==-1||j==-1||k==-1)
return 0;
if(dp[i][j][k]!=-1)
return dp[i][j][k];
if(X[i]==Y[j] && Y[j]==Z[k])
return dp[i][j][k] = 1+lcsOf3(i-1,j-1,k-1);
else
return dp[i][j][k] = max(max(lcsOf3(i-1,j,k),
lcsOf3(i,j-1,k)),lcsOf3(i,j,k-1));
}
// Driver code
int main()
{
memset(dp, -1,sizeof(dp));
int m = X.length();
int n = Y.length();
int o = Z.length();
cout << "Length of LCS is " << lcsOf3(m-1,n-1,o-1);
// this code is contributed by Kushdeep Mittal
}
Java
// Java program to find LCS of three strings
class GFG
{
static String X = "AGGT12";
static String Y = "12TXAYB";
static String Z = "12XBA";
static int[][][] dp = new int[100][100][100];
/* Returns length of LCS for X[0..m-1],
Y[0..n-1] and Z[0..o-1] */
static int lcsOf3(int i, int j, int k)
{
if (i == -1 || j == -1 || k == -1)
{
return 0;
}
if (dp[i][j][k] != -1)
{
return dp[i][j][k];
}
if (X.charAt(i) == Y.charAt(j) &&
Y.charAt(j) == Z.charAt(k))
{
return dp[i][j][k] = 1 + lcsOf3(i - 1, j - 1, k - 1);
} else {
return dp[i][j][k] = Math.max(Math.max(lcsOf3(i - 1, j, k),
lcsOf3(i, j - 1, k)),
lcsOf3(i, j, k - 1));
}
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 100; i++)
{
for (int j = 0; j < 100; j++)
{
for (int k = 0; k < 100; k++)
{
dp[i][j][k] = -1;
}
}
}
int m = X.length();
int n = Y.length();
int o = Z.length();
System.out.print("Length of LCS is "
+ lcsOf3(m - 1, n - 1, o - 1));
}
}
// This code has been contributed by 29AjayKumar
Python3
# Python3 program to find LCS of
# three strings
X = "AGGT12"
Y = "12TXAYB"
Z = "12XBA"
dp = [[[-1 for i in range(100)]
for j in range(100)]
for k in range(100)]
# Returns length of LCS for
# X[0..m-1], Y[0..n-1] and Z[0..o-1]
def lcsOf3(i, j, k) :
if(i == -1 or j == -1 or k == -1) :
return 0
if(dp[i][j][k] != -1) :
return dp[i][j][k]
if(X[i] == Y[j] and Y[j] == Z[k]) :
dp[i][j][k] = 1 + lcsOf3(i - 1,
j - 1, k - 1)
return dp[i][j][k]
else :
dp[i][j][k] = max(max(lcsOf3(i - 1, j, k),
lcsOf3(i, j - 1, k)),
lcsOf3(i, j, k - 1))
return dp[i][j][k]
# Driver code
if __name__ == "__main__" :
m = len(X)
n = len(Y)
o = len(Z)
print("Length of LCS is",
lcsOf3(m - 1, n - 1, o - 1))
# This code is contributed by Ryuga
C#
// C# program to find LCS of three strings
using System;
class GFG
{
static string X = "AGGT12";
static string Y = "12TXAYB";
static string Z = "12XBA";
static int[,,] dp = new int[100, 100, 100];
/* Returns length of LCS for X[0..m-1],
Y[0..n-1] and Z[0..o-1] */
static int lcsOf3(int i, int j, int k)
{
if(i == -1 || j == -1 || k == -1)
return 0;
if(dp[i, j, k] != -1)
return dp[i, j, k];
if(X[i] == Y[j] && Y[j] == Z[k])
return dp[i, j, k] = 1 + lcsOf3(i - 1, j - 1, k - 1);
else
return dp[i, j, k] = Math.Max(Math.Max(lcsOf3(i - 1, j, k),
lcsOf3(i, j - 1, k)),
lcsOf3(i, j, k - 1));
}
// Driver code
static void Main()
{
for(int i = 0; i < 100; i++)
for(int j = 0; j < 100; j++)
for(int k = 0; k < 100; k++)
dp[i, j, k] = -1;
int m = X.Length;
int n = Y.Length;
int o = Z.Length;
Console.Write("Length of LCS is " +
lcsOf3(m - 1, n - 1, o - 1));
}
}
// This code is contributed by DrRoot_
PHP
<?php
// PHP program to find LCS of three strings
$X = "AGGT12";
$Y = "12TXAYB";
$Z = "12XBA";
$dp=array_fill(0, 100, array_fill(0, 100, array_fill(0, 100, -1)));
/* Returns length of LCS for X[0..m-1], Y[0..n-1]
and Z[0..o-1] */
function lcsOf3($i, $j, $k)
{
global $dp, $X, $Y, $Z;
if($i == -1 || $j == -1 || $k == -1)
return 0;
if($dp[$i][$j][$k] != -1)
return $dp[$i][$j][$k];
if($X[$i] == $Y[$j] && $Y[$j] == $Z[$k])
return $dp[$i][$j][$k] = 1+lcsOf3($i - 1, $j - 1, $k - 1);
else
return $dp[$i][$j][$k] = max(max(lcsOf3($i - 1, $j, $k),
lcsOf3($i, $j - 1, $k)), lcsOf3($i, $j, $k - 1));
}
// Driver code
$m = strlen($X);
$n = strlen($Y);
$o = strlen($Z);
echo "Length of LCS is ".lcsOf3($m - 1, $n - 1, $o - 1);
// This code is contributed by mits
?>
Javascript
<script>
// Java program to find LCS of three strings
let X = "AGGT12";
let Y = "12TXAYB";
let Z = "12XBA";
let dp = new Array(100);
for(let i=0;i<100;i++)
{
dp[i]=new Array(100);
for(let j=0;j<100;j++)
{
dp[i][j]=new Array(100);
for(let k=0;k<100;k++)
{
dp[i][j][k]=-1;
}
}
}
/* Returns length of LCS for X[0..m-1],
Y[0..n-1] and Z[0..o-1] */
function lcsOf3(i,j,k)
{
if (i == -1 || j == -1 || k == -1)
{
return 0;
}
if (dp[i][j][k] != -1)
{
return dp[i][j][k];
}
if (X[i] == Y[j] &&
Y[j] == Z[k])
{
return dp[i][j][k] = 1 + lcsOf3(i - 1, j - 1, k - 1);
} else {
return dp[i][j][k] = Math.max(Math.max(lcsOf3(i - 1, j, k),
lcsOf3(i, j - 1, k)),
lcsOf3(i, j, k - 1));
}
}
// Driver code
let m = X.length;
let n = Y.length;
let o = Z.length;
document.write("Length of LCS is "
+ lcsOf3(m - 1, n - 1, o - 1));
// This code is contributed by rag2127
</script>
Producción
Length of LCS is 2
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA