Dada la string str , la tarea es encontrar la cantidad mínima de caracteres que se insertarán para convertirla en un palíndromo.
Antes de continuar, entendamos con algunos ejemplos:
- ab: el número de inserciones necesarias es 1, es decir, b ab
- aa: el número de inserciones requeridas es 0, es decir, aa
- abcd: el número de inserciones necesarias es 3, es decir , dcb abcd
- abcda: El número de inserciones requeridas es 2, es decir, un dc bcda que es el mismo que el número de inserciones en la substring bcd (¿Por qué?).
- abcde: el número de inserciones necesarias es 4, es decir , edcb abcde
Deje que la string de entrada sea str[l……h] . El problema se puede dividir en tres partes:
- Encuentre el número mínimo de inserciones en la substring str[l+1,…….h].
- Encuentre el número mínimo de inserciones en la substring str[l…….h-1].
- Encuentre el número mínimo de inserciones en la substring str[l+1……h-1].
Enfoque recursivo : el número mínimo de inserciones en la string str[l…..h] se puede dar como:
- minInsertions(str[l+1…..h-1]) si str[l] es igual a str[h]
- min(minInsertions(str[l…..h-1]), minInsertions(str[l+1…..h])) + 1 en caso contrario
A continuación se muestra la implementación del enfoque anterior:
C++
// A Naive recursive program to find minimum
// number insertions needed to make a string
// palindrome
#include<bits/stdc++.h>
using namespace std;
// Recursive function to find
// minimum number of insertions
int findMinInsertions(char str[], int l, int h)
{
// Base Cases
if (l > h) return INT_MAX;
if (l == h) return 0;
if (l == h - 1) return (str[l] == str[h])? 0 : 1;
// Check if the first and last characters are
// same. On the basis of the comparison result,
// decide which subproblem(s) to call
return (str[l] == str[h])?
findMinInsertions(str, l + 1, h - 1):
(min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1);
}
// Driver code
int main()
{
char str[] = "geeks";
cout << findMinInsertions(str, 0, strlen(str) - 1);
return 0;
}
// This code is contributed by
// Akanksha Rai
C
// A Naive recursive program to find minimum
// number insertions needed to make a string
// palindrome
#include <stdio.h>
#include <limits.h>
#include <string.h>
// A utility function to find minimum of two numbers
int min(int a, int b)
{ return a < b ? a : b; }
// Recursive function to find minimum number of
// insertions
int findMinInsertions(char str[], int l, int h)
{
// Base Cases
if (l > h) return INT_MAX;
if (l == h) return 0;
if (l == h - 1) return (str[l] == str[h])? 0 : 1;
// Check if the first and last characters are
// same. On the basis of the comparison result,
// decide which subproblem(s) to call
return (str[l] == str[h])?
findMinInsertions(str, l + 1, h - 1):
(min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1);
}
// Driver program to test above functions
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertions(str, 0, strlen(str)-1));
return 0;
}
Java
// A Naive recursive Java program to find minimum
// number insertions needed to make a string
// palindrome
class GFG {
// Recursive function to find minimum number
// of insertions
static int findMinInsertions(char str[], int l,
int h)
{
// Base Cases
if (l > h) return Integer.MAX_VALUE;
if (l == h) return 0;
if (l == h - 1) return (str[l] == str[h])? 0 : 1;
// Check if the first and last characters
// are same. On the basis of the comparison
// result, decide which subproblem(s) to call
return (str[l] == str[h])?
findMinInsertions(str, l + 1, h - 1):
(Integer.min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1);
}
// Driver program to test above functions
public static void main(String args[])
{
String str= "geeks";
System.out.println(findMinInsertions(str.toCharArray(),
0, str.length()-1));
}
}
// This code is contributed by Sumit Ghosh
Python 3
# A Naive recursive program to find minimum # number insertions needed to make a string # palindrome import sys # Recursive function to find minimum # number of insertions def findMinInsertions(str, l, h): # Base Cases if (l > h): return sys.maxsize if (l == h): return 0 if (l == h - 1): return 0 if(str[l] == str[h]) else 1 # Check if the first and last characters are # same. On the basis of the comparison result, # decide which subproblem(s) to call if(str[l] == str[h]): return findMinInsertions(str, l + 1, h - 1) else: return (min(findMinInsertions(str, l, h - 1), findMinInsertions(str, l + 1, h)) + 1) # Driver Code if __name__ == "__main__": str = "geeks" print(findMinInsertions(str, 0, len(str) - 1)) # This code is contributed by ita_c
C#
// A Naive recursive C# program
// to find minimum number
// insertions needed to make
// a string palindrome
using System;
class GFG
{
// Recursive function to
// find minimum number of
// insertions
static int findMinInsertions(char []str,
int l, int h)
{
// Base Cases
if (l > h) return int.MaxValue;
if (l == h) return 0;
if (l == h - 1)
return (str[l] == str[h])? 0 : 1;
// Check if the first and
// last characters are same.
// On the basis of the
// comparison result, decide
// which subproblem(s) to call
return (str[l] == str[h])?
findMinInsertions(str,
l + 1, h - 1):
(Math.Min(findMinInsertions(str, l,
h - 1),
findMinInsertions(str, l +
1, h)) + 1);
}
// Driver Code
public static void Main()
{
string str= "geeks";
Console.WriteLine(findMinInsertions(str.ToCharArray(),
0, str.Length - 1));
}
}
// This code is contributed by Sam007
Javascript
<script>
// A Naive recursive JavaScript program to find minimum
// number insertions needed to make a string
// palindrome
// Recursive function to find minimum number
// of insertions
function findMinInsertions(str,l,h)
{
// Base Cases
if (l > h)
return Number.MAX_VALUE;
if (l == h)
return 0;
if (l == h - 1)
return (str[l] == str[h])? 0 : 1;
// Check if the first and last characters
// are same. On the basis of the comparison
// result, decide which subproblem(s) to call
return (str[l] == str[h]) ?
findMinInsertions(str, l + 1, h - 1) :
(Math.min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1)
}
// Driver program to test above functions
let str= "geeks";
document.write(findMinInsertions(str,0, str.length-1));
// This code is contributed by rag2127
</script>
Producción
3
Tiempo Complejidad : O(n n )
Espacio Auxiliar: O(n)
Solución basada en programación dinámica
Si observamos cuidadosamente el enfoque anterior, podemos encontrar que presenta subproblemas superpuestos .
Supongamos que queremos encontrar el número mínimo de inserciones en la string «abcde»:
abcde
/ | \
/ | \
bcde abcd bcd <- case 3 is discarded as str[l] != str[h]
/ | \ / | \
/ | \ / | \
cde bcd cd bcd abc bc
/ | \ / | \ /|\ / | \
de cd d cd bc c………………….
Las substrings en negrita muestran que la recursividad debe terminar y el árbol de recurrencia no puede originarse desde allí. La substring del mismo color indica subproblemas superpuestos .
¿Cómo reutilizar soluciones de subproblemas? La técnica de memorización se utiliza para evitar recordar subproblemas similares. Podemos crear una tabla para almacenar los resultados de los subproblemas para que puedan usarse directamente si se vuelve a encontrar el mismo subproblema.
La siguiente tabla representa los valores almacenados para la string abcde.
a b c d e ---------- 0 1 2 3 4 0 0 1 2 3 0 0 0 1 2 0 0 0 0 1 0 0 0 0 0
¿Cómo llenar la mesa?
La tabla debe llenarse en forma diagonal. Para la string abcde, 0….4, se debe ordenar lo siguiente en que se llena la tabla:
Gap = 1: (0, 1) (1, 2) (2, 3) (3, 4) Gap = 2: (0, 2) (1, 3) (2, 4) Gap = 3: (0, 3) (1, 4) Gap = 4: (0, 4)
A continuación se muestra la implementación del enfoque anterior:
C++
// A Dynamic Programming based program to find
// minimum number insertions needed to make a
// string palindrome
#include <bits/stdc++.h>
using namespace std;
// A DP function to find minimum
// number of insertions
int findMinInsertionsDP(char str[], int n)
{
// Create a table of size n*n. table[i][j]
// will store minimum number of insertions
// needed to convert str[i..j] to a palindrome.
int table[n][n], l, h, gap;
// Initialize all table entries as 0
memset(table, 0, sizeof(table));
// Fill the table
for (gap = 1; gap < n; ++gap)
for (l = 0, h = gap; h < n; ++l, ++h)
table[l][h] = (str[l] == str[h])?
table[l + 1][h - 1] :
(min(table[l][h - 1],
table[l + 1][h]) + 1);
// Return minimum number of insertions
// for str[0..n-1]
return table[0][n - 1];
}
// Driver Code
int main()
{
char str[] = "geeks";
cout << findMinInsertionsDP(str, strlen(str));
return 0;
}
// This is code is contributed by rathbhupendra
C
// A Dynamic Programming based program to find
// minimum number insertions needed to make a
// string palindrome
#include <stdio.h>
#include <string.h>
// A utility function to find minimum of two integers
int min(int a, int b)
{ return a < b ? a : b; }
// A DP function to find minimum number of insertions
int findMinInsertionsDP(char str[], int n)
{
// Create a table of size n*n. table[i][j]
// will store minimum number of insertions
// needed to convert str[i..j] to a palindrome.
int table[n][n], l, h, gap;
// Initialize all table entries as 0
memset(table, 0, sizeof(table));
// Fill the table
for (gap = 1; gap < n; ++gap)
for (l = 0, h = gap; h < n; ++l, ++h)
table[l][h] = (str[l] == str[h])?
table[l+1][h-1] :
(min(table[l][h-1],
table[l+1][h]) + 1);
// Return minimum number of insertions for
// str[0..n-1]
return table[0][n-1];
}
// Driver program to test above function.
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertionsDP(str, strlen(str)));
return 0;
}
Java
// A Java solution for Dynamic Programming
// based program to find minimum number
// insertions needed to make a string
// palindrome
import java.util.Arrays;
class GFG
{
// A DP function to find minimum number
// of insertions
static int findMinInsertionsDP(char str[], int n)
{
// Create a table of size n*n. table[i][j]
// will store minimum number of insertions
// needed to convert str[i..j] to a palindrome.
int table[][] = new int[n][n];
int l, h, gap;
// Fill the table
for (gap = 1; gap < n; ++gap)
for (l = 0, h = gap; h < n; ++l, ++h)
table[l][h] = (str[l] == str[h])?
table[l+1][h-1] :
(Integer.min(table[l][h-1],
table[l+1][h]) + 1);
// Return minimum number of insertions
// for str[0..n-1]
return table[0][n-1];
}
// Driver program to test above function.
public static void main(String args[])
{
String str = "geeks";
System.out.println(
findMinInsertionsDP(str.toCharArray(), str.length()));
}
}
// This code is contributed by Sumit Ghosh
Python3
# A Dynamic Programming based program to # find minimum number insertions needed # to make a string palindrome # A utility function to find minimum # of two integers def Min(a, b): return min(a, b) # A DP function to find minimum number # of insertions def findMinInsertionsDP(str1, n): # Create a table of size n*n. table[i][j] # will store minimum number of insertions # needed to convert str1[i..j] to a palindrome. table = [[0 for i in range(n)] for i in range(n)] l, h, gap = 0, 0, 0 # Fill the table for gap in range(1, n): l = 0 for h in range(gap, n): if str1[l] == str1[h]: table[l][h] = table[l + 1][h - 1] else: table[l][h] = (Min(table[l][h - 1], table[l + 1][h]) + 1) l += 1 # Return minimum number of insertions # for str1[0..n-1] return table[0][n - 1]; # Driver Code str1 = "geeks" print(findMinInsertionsDP(str1, len(str1))) # This code is contributed by # Mohit kumar 29
C#
// A C# solution for Dynamic Programming
// based program to find minimum number
// insertions needed to make a string
// palindrome
using System;
class GFG
{
// A DP function to find minimum number
// of insertions
static int findMinInsertionsDP(char []str, int n)
{
// Create a table of size n*n. table[i][j]
// will store minimum number of insertions
// needed to convert str[i..j] to a palindrome.
int [,]table = new int[n, n];
int l, h, gap;
// Fill the table
for (gap = 1; gap < n; ++gap)
for (l = 0, h = gap; h < n; ++l, ++h)
table[l, h] = (str[l] == str[h])?
table[l+1, h-1] :
(Math.Min(table[l, h-1],
table[l+1, h]) + 1);
// Return minimum number of insertions
// for str[0..n-1]
return table[0, n-1];
}
// Driver code
public static void Main()
{
String str = "geeks";
Console.Write(
findMinInsertionsDP(str.ToCharArray(), str.Length));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// A Javascript solution for Dynamic Programming
// based program to find minimum number
// insertions needed to make a string
// palindrome
// A DP function to find minimum number
// of insertions
function findMinInsertionsDP(str,n)
{
// Create a table of size n*n. table[i][j]
// will store minimum number of insertions
// needed to convert str[i..j] to a palindrome.
let table=new Array(n);
for(let i=0;i<n;i++)
{
table[i]=new Array(n);
}
for(let i=0;i<n;i++)
{
for(let j=0;j<n;j++)
{
table[i][j]=0;
}
}
let l=0, h=0, gap=0;
// Fill the table
for (gap = 1; gap < n; gap++)
{
for (l = 0, h = gap; h < n; l++, h++)
{
table[l][h] = (str[l] == str[h]) ? table[l+1][h-1] : (Math.min(table[l][h-1],table[l+1][h]) + 1);
}
}
// Return minimum number of insertions
// for str[0..n-1]
return table[0][n - 1];
}
// Driver program to test above function.
let str = "geeks";
document.write(findMinInsertionsDP(str, str.length));
// This code is contributed by avanitrachhadiya2155
</script>
Producción
3
Complejidad temporal: O(N 2 )
Espacio auxiliar: O(N 2 )
Otra solución de programación dinámica (variación del problema de la subsecuencia común más larga)
El problema de encontrar inserciones mínimas también se puede resolver utilizando el problema de la subsecuencia común más larga (LCS) . Si averiguamos el LCS de string y su reverso, sabemos cuántos caracteres como máximo pueden formar un palíndromo. Necesitamos insertar los caracteres restantes. Los siguientes son los pasos.
- Encuentre la longitud de LCS de la string de entrada y su reverso. Sea la longitud ‘l’.
- El número mínimo de inserciones necesarias es la longitud de la string de entrada menos ‘l’.
A continuación se muestra la implementación del enfoque anterior:
C++
// An LCS based program to find minimum number
// insertions needed to make a string palindrome
#include <bits/stdc++.h>
using namespace std;
// Returns length of LCS for X[0..m-1], Y[0..n-1].
int lcs( string X, string Y, int m, int n )
{
int L[m+1][n+1];
int i, j;
/* Following steps build L[m+1][n+1] in bottom
up fashion. Note that L[i][j] contains length
of LCS of X[0..i-1] and Y[0..j-1] */
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = max(L[i - 1][j], L[i][j - 1]);
}
}
/* L[m][n] contains length of LCS for X[0..n-1]
and Y[0..m-1] */
return L[m][n];
}
void reverseStr(string& str)
{
int n = str.length();
// Swap character starting from two
// corners
for (int i = 0; i < n / 2; i++)
swap(str[i], str[n - i - 1]);
}
// LCS based function to find minimum number of
// insertions
int findMinInsertionsLCS(string str, int n)
{
// Creata another string to store reverse of 'str'
string rev = "";
rev = str;
reverseStr(rev);
// The output is length of string minus length of lcs of
// str and it reverse
return (n - lcs(str, rev, n, n));
}
// Driver code
int main()
{
string str = "geeks";
cout << findMinInsertionsLCS(str, str.length());
return 0;
}
// This code is contributed by rathbhupendra
C
// An LCS based program to find minimum number
// insertions needed to make a string palindrome
#include<stdio.h>
#include <string.h>
/* Utility function to get max of 2 integers */
int max(int a, int b)
{ return (a > b)? a : b; }
/* Returns length of LCS for X[0..m-1], Y[0..n-1].
See http://goo.gl/bHQVP for details of this
function */
int lcs( char *X, char *Y, int m, int n )
{
int L[m+1][n+1];
int i, j;
/* Following steps build L[m+1][n+1] in bottom
up fashion. Note that L[i][j] contains length
of LCS of X[0..i-1] and Y[0..j-1] */
for (i=0; i<=m; i++)
{
for (j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i-1] == Y[j-1])
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = max(L[i-1][j], L[i][j-1]);
}
}
/* L[m][n] contains length of LCS for X[0..n-1]
and Y[0..m-1] */
return L[m][n];
}
// LCS based function to find minimum number of
// insertions
int findMinInsertionsLCS(char str[], int n)
{
// Creata another string to store reverse of 'str'
char rev[n+1];
strcpy(rev, str);
strrev(rev);
// The output is length of string minus length of lcs of
// str and it reverse
return (n - lcs(str, rev, n, n));
}
// Driver program to test above functions
int main()
{
char str[] = "geeks";
printf("%d", findMinInsertionsLCS(str, strlen(str)));
return 0;
}
Java
// An LCS based Java program to find minimum
// number insertions needed to make a string
// palindrome
class GFG
{
/* Returns length of LCS for X[0..m-1],
Y[0..n-1]. See http://goo.gl/bHQVP for
details of this function */
static int lcs( String X, String Y, int m, int n )
{
int L[][] = new int[m+1][n+1];
int i, j;
/* Following steps build L[m+1][n+1] in
bottom up fashion. Note that L[i][j]
contains length of LCS of X[0..i-1]
and Y[0..j-1] */
for (i=0; i<=m; i++)
{
for (j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i-1) == Y.charAt(j-1))
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = Integer.max(L[i-1][j], L[i][j-1]);
}
}
/* L[m][n] contains length of LCS for
X[0..n-1] and Y[0..m-1] */
return L[m][n];
}
// LCS based function to find minimum number
// of insertions
static int findMinInsertionsLCS(String str, int n)
{
// Using StringBuffer to reverse a String
StringBuffer sb = new StringBuffer(str);
sb.reverse();
String revString = sb.toString();
// The output is length of string minus
// length of lcs of str and it reverse
return (n - lcs(str, revString , n, n));
}
// Driver program to test above functions
public static void main(String args[])
{
String str = "geeks";
System.out.println(
findMinInsertionsLCS(str, str.length()));
}
}
// This code is contributed by Sumit Ghosh
Python3
# An LCS based Python3 program to find minimum # number insertions needed to make a string # palindrome """ Returns length of LCS for X[0..m-1], Y[0..n-1]. See http://goo.gl/bHQVP for details of this function """ def lcs(X, Y, m, n) : L = [[0 for i in range(n + 1)] for j in range(m + 1)] """ Following steps build L[m + 1, n + 1] in bottom up fashion. Note that L[i, j] contains length of LCS of X[0..i - 1] and Y[0..j - 1] """ for i in range(m + 1) : for j in range(n + 1) : if (i == 0 or j == 0) : L[i][j] = 0 elif (X[i - 1] == Y[j - 1]) : L[i][j] = L[i - 1][j - 1] + 1 else : L[i][j] = max(L[i - 1][j], L[i][j - 1]) """ L[m,n] contains length of LCS for X[0..n-1] and Y[0..m-1] """ return L[m][n] # LCS based function to find minimum number # of insertions def findMinInsertionsLCS(Str, n) : # Using charArray to reverse a String charArray = list(Str) charArray.reverse() revString = "".join(charArray) # The output is length of string minus # length of lcs of str and it reverse return (n - lcs(Str, revString , n, n)) # Driver code Str = "geeks" print(findMinInsertionsLCS(Str,len(Str))) # This code is contributed by divyehrabadiya07
C#
// An LCS based C# program to find minimum
// number insertions needed to make a string
// palindrome
using System;
class GFG
{
/* Returns length of LCS for X[0..m-1],
Y[0..n-1]. See http://goo.gl/bHQVP for
details of this function */
static int lcs( string X, string Y, int m, int n )
{
int[,] L = new int[m + 1, n + 1];
int i, j;
/* Following steps build L[m+1,n+1] in
bottom up fashion. Note that L[i,j]
contains length of LCS of X[0..i-1]
and Y[0..j-1] */
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i, j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i, j] = L[i - 1, j - 1] + 1;
else
L[i, j] = Math.Max(L[i - 1, j], L[i, j - 1]);
}
}
/* L[m,n] contains length of LCS for
X[0..n-1] and Y[0..m-1] */
return L[m,n];
}
// LCS based function to find minimum number
// of insertions
static int findMinInsertionsLCS(string str, int n)
{
// Using charArray to reverse a String
char[] charArray = str.ToCharArray();
Array.Reverse(charArray);
string revString = new string(charArray);
// The output is length of string minus
// length of lcs of str and it reverse
return (n - lcs(str, revString , n, n));
}
// Driver code
static void Main()
{
string str = "geeks";
Console.WriteLine(findMinInsertionsLCS(str,str.Length));
}
}
// This code is contributed by mits
Javascript
<script>
// An LCS based Javascript program to find minimum
// number insertions needed to make a string
// palindrome
/* Returns length of LCS for X[0..m-1],
Y[0..n-1]. See http://goo.gl/bHQVP for
details of this function */
function lcs(X, Y, m, n)
{
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] = 0;
}
}
let i, j;
/* Following steps build L[m+1][n+1] in
bottom up fashion. Note that L[i][j]
contains length of LCS of X[0..i-1]
and Y[0..j-1] */
for (i = 0; i <= m; i++)
{
for (j = 0; j <= n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]);
}
}
/* L[m][n] contains length of LCS for
X[0..n-1] and Y[0..m-1] */
return L[m][n];
}
// LCS based function to find minimum number
// of insertions
function findMinInsertionsLCS(str, n)
{
let revString = str.split('').reverse().join('');
// The output is length of string minus
// length of lcs of str and it reverse
return (n - lcs(str, revString , n, n));
}
// Driver program to test above functions
let str = "geeks";
document.write(findMinInsertionsLCS(str, str.length));
// This code is contributed by unknown2108
</script>
Producción
3
Complejidad temporal: O(N 2 )
Espacio auxiliar : O(N 2 )
Método de optimización de espacio: el código anterior se puede optimizar en el espacio usando solo una array 1d en lugar de una array 2d. En la tabla dp solo necesitamos la fila anterior y los elementos de la fila actual.
C++
// An LCS based program to find minimum number
// insertions needed to make a string palindrome
#include <bits/stdc++.h>
using namespace std;
// Returns length of LCS for X[0..m-1], Y[0..n-1].
int lcs(string X, string Y, int m, int n)
{
vector<int> prev(n + 1, 0), curr(n + 1, 0);
int i, j;
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
prev[j] = 0;
else if (X[i - 1] == Y[j - 1])
curr[j] = prev[j - 1] + 1;
else
curr[j] = max(prev[j], curr[j - 1]);
}
prev = curr;
}
/* L[m][n] contains length of LCS for X[0..n-1]
and Y[0..m-1] */
return prev[n];
}
void reverseStr(string& str)
{
int n = str.length();
// Swap character starting from two
// corners
for (int i = 0; i < n / 2; i++)
swap(str[i], str[n - i - 1]);
}
// LCS based function to find minimum number of
// insertions
int findMinInsertionsLCS(string str, int n)
{
// Creata another string to store reverse of 'str'
string rev = "";
rev = str;
reverseStr(rev);
// The output is length of string minus length of lcs of
// str and it reverse
return (n - lcs(str, rev, n, n));
}
// Driver code
int main()
{
string str = "geeks";
cout << findMinInsertionsLCS(str, str.length());
return 0;
}
// This code is contributed by Sanskar
C#
/* C# program to implement an LCS based approach
to find minimum number of insertions needed
to make a string palindrome*/
using System;
using System.Collections.Generic;
class GFG {
// Returns length of LCS for X[0..m-1], Y[0..n-1].
static int Lcs(string X, string Y, int m, int n)
{
int[] prev = new int[n + 1];
int[] curr = new int[n + 1];
int i, j;
for (i = 0; i <= m; i++) {
for (j = 0; j <= n; j++) {
if (i == 0 || j == 0)
prev[j] = 0;
else if (X[i - 1] == Y[j - 1])
curr[j] = prev[j - 1] + 1;
else
curr[j]
= Math.Max(prev[j], curr[j - 1]);
}
prev = curr;
}
/* L[m][n] contains length of LCS for X[0..n-1]
and Y[0..m-1] */
return prev[n];
}
// LCS based function to find minimum number of
// insertions
static int FindMinInsertionsLCS(string str, int n)
{
// Creat another string to store reverse of 'str'
char[] tmp = str.ToCharArray();
Array.Reverse(tmp);
string rev = new string(tmp);
// The output is length of string minus length of
// lcs of str and it reverse
return (n - Lcs(str, rev, n, n));
}
// Driver code
static void Main(string[] args)
{
string str = "geeks";
Console.WriteLine(
FindMinInsertionsLCS(str, str.Length));
}
}
// This code is contributed by cavi4762
Producción
3
Complejidad temporal: O(N 2 )
Espacio auxiliar: O(N)
Artículo relacionado:
Número mínimo de apéndices necesarios para hacer un palíndromo de strings
Este artículo fue compilado por Aarti_Rathi y Aashish Barnwal . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA