Dadas dos strings A y B de tamaño N y M respectivamente, donde B es una subsecuencia de A y una array arr[] de tamaño N , donde arr[i] es el costo de eliminar el i-ésimo carácter de la string A. La tarea es encontrar el costo mínimo para eliminar caracteres de A de modo que después de la eliminación ninguna subsecuencia de A sea igual a B .
Ejemplos:
Entrada: A = “abccd”, B = “ccd”, arr[] = {1, 2, 3, 4, 5}
Salida: 3
Explicación: si eliminamos la ‘d’ o una sola ‘c’ de A, entonces no será posible construir una subsecuencia igual a B.
Entre estos, el costo de eliminar la ‘c’ en el índice 2 es mínimo, es decir, 3. Entonces, la respuesta es 3.Entrada: A = “abccdabccdabccd”, B = “bccd”, arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Salida : 21
Explicación: Si quitamos las tres ‘b’ de coste 2, 7 y 5, entonces A se convierte en “accdaccdaccd”.
Ahora no hay forma de construir una subsecuencia = «bccd»
Enfoque: El enfoque ingenuo para resolver el problema es usar la recursividad y encontrar una subsecuencia común entre A y B según la siguiente idea:
Si el carácter de A y B coincide, habrá 2 opciones: eliminar ese carácter de A o mantener ese carácter y eliminar otros caracteres.
Siga los pasos mencionados a continuación para implementar la idea:
- Atraviese recursivamente la string A usando el puntero i y mantenga un puntero j para apuntar a B .
- En cada función recursiva:
- Si se llega al final de la string, devuelve 0 si se elimina algún carácter (verificado por un contador de elementos eliminados); de lo contrario, devuelve un valor positivo alto.
- Si A[i] = B[j] hay dos opciones:
- Elimine A[i] y agregue cost arr[i] para responder y llamar recursivamente a los siguientes elementos.
- Mantén A[i] y avanza.
- De lo contrario, sigue avanzando hasta que A[i] coincida con B[i] .
- Devuelve el costo mínimo entre los dos casos anteriores de cada llamada recursiva.
- Devuelve el costo mínimo como respuesta.
Tiempo Complejidad: O(2 M )
Espacio Auxiliar: O(1)
Enfoque eficiente: el tiempo del enfoque anterior se puede optimizar aún más utilizando la programación dinámica según la siguiente idea:
Use una array 3D dp[][][] para almacenar el costo mínimo hasta una posición determinada en A y B y para eliminar al menos un carácter. dp[i][j][] almacena el costo mínimo para eliminar caracteres hasta el i-ésimo índice de A y el j-ésimo índice de B donde al menos un carácter se elimina o no.
Por lo tanto, la tercera dimensión solo tiene dos valores, ya sea 1 (se elimina al menos uno) o 0 (no se elimina ninguno)
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Array for memoization
int dp[101][101][2];
// Recursive function to calculate
// the minimum cost using dynamic programming
int minCostUtil(string& a, int n,
string& b, int m,
vector<int>& c, int removed)
{
// Base case reached the end of string
if (n == 0 || m == 0) {
// Removed 0 characters
// return high (+)ve value
if (removed == 0)
return 99999;
return 0;
}
// Return pre-calculated value
if (dp[n][m][removed > 0 ? 1 : 0] != -1)
return dp[n][m][removed > 0 ? 1 : 0];
// If characters match return the minimum of
// 1. Removing the character from A and
// adding the cost
// 2. Moving forward to remove some other
// character and decrease the counter as
// this character will not be removed.
if (a[n - 1] == b[m - 1]) {
dp[n][m][removed > 0 ? 1 : 0]
= min(c[n - 1]
+ minCostUtil(a, n - 1,
b, m, c, removed),
minCostUtil(a, n - 1, b, m - 1,
c, removed - 1));
return dp[n][m][removed > 0 ? 1 : 0];
}
// If no match then move through string
// A and try removing some other
// character which matches, i.e can be
// part of the subsequence that is equal to B
else
return dp[n][m][removed > 0 ? 1 : 0]
= minCostUtil(a, n - 1,
b, m, c, removed);
}
// Function to calculate minimum cost
int minCost(string& a, string& b,
vector<int>& c)
{
memset(dp, -1, sizeof(dp));
return minCostUtil(a, a.size(), b,
b.size(), c, b.size());
}
// Driver code
int main()
{
string A = "abccdabccdabccd";
string B = "bccd";
vector<int> arr = { 1, 2, 3, 4, 5,
6, 7, 8, 9, 10,
11, 12, 13, 14, 15 };
cout << minCost(A, B, arr);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
public class GFG {
// Array for memoization
static int dp[][][] = new int[101][101][2];
// Recursive function to calculate
// the minimum cost using dynamic programming
static int minCostUtil(String a, int n, String b, int m,
int[] c, int removed)
{
// Base case reached the end of string
if (n == 0 || m == 0) {
// Removed 0 characters
// return high (+)ve value
if (removed == 0)
return 99999;
return 0;
}
// Return pre-calculated value
if (dp[n][m][removed > 0 ? 1 : 0] != -1)
return dp[n][m][removed > 0 ? 1 : 0];
// If characters match return the minimum of
// 1. Removing the character from A and
// adding the cost
// 2. Moving forward to remove some other
// character and decrease the counter as
// this character will not be removed.
if (a.charAt(n - 1) == b.charAt(m - 1)) {
dp[n][m][removed > 0 ? 1 : 0]
= Math.min(c[n - 1]
+ minCostUtil(a, n - 1, b, m,
c, removed),
minCostUtil(a, n - 1, b, m - 1,
c, removed - 1));
return dp[n][m][removed > 0 ? 1 : 0];
}
// If no match then move through string
// A and try removing some other
// character which matches, i.e can be
// part of the subsequence that is equal to B
else
return dp[n][m][removed > 0 ? 1 : 0]
= minCostUtil(a, n - 1, b, m, c, removed);
}
// Function to calculate minimum cost
static int minCost(String a, String b, int[] c)
{
for (int i = 0; i < 101; i++) {
for (int j = 0; j < 101; j++) {
for (int k = 0; k < 2; k++) {
dp[i][j][k] = -1;
}
}
}
return minCostUtil(a, a.length(), b, b.length(), c,
b.length());
}
// Driver code
public static void main(String args[])
{
String A = "abccdabccdabccd";
String B = "bccd";
int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15 };
System.out.print(minCost(A, B, arr));
}
}
// This code is contributed by Samim Hossain Mondal.
Python3
# Python3 program for the above approach # Array for memoization dp = [] # recursive function to calculate the # minimum cost using dynamic programming def minCostUtil(a, n, b, m, c, removed): global dp # Base Case - reached the end of the string if n == 0 or m == 0: # removed 0 characters # return high +ve value if removed == 0: return 99999 return 0 # return pre - calculated value if dp[n][m][int(bool(removed))] != -1: return dp[n][m][int(bool(removed))] # 1. Removing the character from A and # adding the cost # 2. Moving forward to remove some other # character and decrease the counter as # this character will not be removed. if a[n - 1] == b[m - 1]: dp[n][m][int(bool(removed))] = min(c[n - 1] + minCostUtil(a, n - 1, b, m, c, removed), minCostUtil(a, n - 1, b, m - 1, c, removed - 1)) return dp[n][m][int(bool(removed))] # if no match, then move through string # A and try removing some other # character which matches, ie, can be # part of the subsequence that is equal to B else: dp[n][m][int(bool(removed))] = minCostUtil(a, n - 1, b, m, c, removed) return dp[n][m][int(bool(removed))] # function to calculate minimum bed def minCost(a, b, c): global dp for i in range(101): dp.append([]) for j in range(101): dp[i].append([]) for k in range(2): dp[i][j].append(-1) return minCostUtil(a, len(a), b, len(b), c, len(b)) # Driver Code A = "abccdabccdabccd" B = "bccd" arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15] print(minCost(A, B, arr)) # This code is contributed by phasing17
C#
// C# program for the above approach
using System;
public class GFG{
// Array for memoization
static int[,,] dp = new int[101, 101, 2];
// Recursive function to calculate
// the minimum cost using dynamic programming
static int minCostUtil(string a, int n, string b, int m,
int[] c, int removed)
{
// Base case reached the end of string
if (n == 0 || m == 0) {
// Removed 0 characters
// return high (+)ve value
if (removed == 0)
return 99999;
return 0;
}
// Return pre-calculated value
if (dp[n, m, removed > 0 ? 1 : 0] != -1)
return dp[n, m, removed > 0 ? 1 : 0];
// If characters match return the minimum of
// 1. Removing the character from A and
// adding the cost
// 2. Moving forward to remove some other
// character and decrease the counter as
// this character will not be removed.
if (a[n - 1] == b[m - 1]) {
dp[n, m, removed > 0 ? 1 : 0]
= Math.Min(c[n - 1]
+ minCostUtil(a, n - 1, b, m,
c, removed),
minCostUtil(a, n - 1, b, m - 1,
c, removed - 1));
return dp[n, m, removed > 0 ? 1 : 0];
}
// If no match then move through string
// A and try removing some other
// character which matches, i.e can be
// part of the subsequence that is equal to B
else
return dp[n, m, removed > 0 ? 1 : 0]
= minCostUtil(a, n - 1, b, m, c, removed);
}
// Function to calculate minimum cost
static int minCost(string a, string b, int[] c)
{
for (int i = 0; i < 101; i++) {
for (int j = 0; j < 101; j++) {
for (int k = 0; k < 2; k++) {
dp[i, j, k] = -1;
}
}
}
return minCostUtil(a, a.Length, b, b.Length, c,
b.Length);
}
// Driver code
static public void Main (){
string A = "abccdabccdabccd";
string B = "bccd";
int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8,
9, 10, 11, 12, 13, 14, 15 };
Console.Write(minCost(A, B, arr));
}
}
// This code is contributed by hrithikgarg03188.
Javascript
// JavaScript program for the above approach
// Array for memoization
var dp = [];
// Recursive function to calculate
// the minimum cost using dynamic programming
function minCostUtil(a, n, b, m, c, removed)
{
// Base case reached the end of string
if (n == 0 || m == 0) {
// Removed 0 characters
// return high (+)ve value
if (removed == 0)
return 99999;
return 0;
}
// Return pre-calculated value
if (dp[n][m][Number(Boolean(removed))] != -1)
return dp[n][m][(removed > 0) ? 1 : 0];
// If characters match return the minimum of
// 1. Removing the character from A and
// adding the cost
// 2. Moving forward to remove some other
// character and decrease the counter as
// this character will not be removed.
if (a[n - 1] == b[m - 1]) {
dp[n][m][(removed > 0) ? 1 : 0] = Math.min(
c[n - 1]
+ minCostUtil(a, n - 1, b, m, c, removed),
minCostUtil(a, n - 1, b, m - 1, c,
removed - 1));
return dp[n][m][(removed > 0) ? 1 : 0];
}
// If no match then move through string
// A and try removing some other
// character which matches, i.e can be
// part of the subsequence that is equal to B
else
return dp[n][m][(removed > 0) ? 1 : 0]
= minCostUtil(a, n - 1, b, m, c, removed);
}
// Function to calculate minimum cost
function minCost(a, b, c)
{
for (var i = 0; i < 101; i++) {
dp[i] = [];
for (var j = 0; j < 101; j++) {
dp[i].push([]);
for (var k = 0; k < 2; k++) {
dp[i][j].push([-1]);
}
}
}
return minCostUtil(a, a.length, b, b.length, c,
b.length);
}
// Driver code
var A = "abccdabccdabccd";
var B = "bccd";
var arr =
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 ];
document.write(minCost(A, B, arr));
// This code is contributed by phasing17
Producción
21
Complejidad temporal: O(N*M)
Espacio auxiliar: O(N*M)