Dadas dos arrays array[] y multiplicadores[] de tamaño N y M donde N siempre es mayor que igual a M. Hay M operaciones a realizar. En cada operación, elija el multiplicador[i] y un elemento de la array arr[], ya sea desde el principio o el final, digamos K , luego agregue el multiplicador[i]*K al puntaje total, digamos, y elimine K de la array arr[ ]. La tarea es encontrar el valor máximo de la puntuación final ans.
Ejemplos:
Entrada : array[] = {1, 2, 3}, multiplicadores[] = {3, 2, 1}, N=3, M=3
Salida : 14
Explicación : una solución óptima es la siguiente:
– Elija del final , [1, 2, 3], sumando 3 * 3 = 9 a la puntuación.
– Elige del final, [1, 2], sumando 2 * 2 = 4 a la puntuación.
– Elija del final, [1], sumando 1 * 1 = 1 a la puntuación.
La puntuación total es 9 + 4 + 1 = 14.Entrada: array[] = {2, 1}, multiplicadores[] = {0}, N=2, M=1
Salida: 0
Explicación: No importa si se elige 2 o 1, la respuesta será 0 porque multiplicador[0] es igual a 0
Enfoque ingenuo: la solución de fuerza bruta es verificar todos y cada uno de los pares recursivamente y encontrar la solución óptima.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum score
// using dynamic programming and
// memoization
int getMaxScore(vector<int>& array,
vector<int>& multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.size(), N = array.size();
int remain = N - M;
vector<int> dp(M + 1, 0);
for (int i = 0; i < M; ++i) {
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j) {
dp[j] = max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
int main()
{
vector<int> array = { 1, 2, 3 };
vector<int> multipliers = { 3, 2, 1 };
cout << getMaxScore(array, multipliers);
return 0;
}
// This code is contributed by rakeshsahni
Java
// Java program for the above approach
public class GFG {
// Function to find the maximum score
// using dynamic programming and
// memoization
static int getMaxScore(int []array,int []multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.length;
int N = array.length;
int remain = N - M;
int dp[] = new int[M + 1];
for (int i = 0; i < M; ++i) {
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j) {
dp[j] = Math.max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
public static void main (String[] args)
{
int []array = { 1, 2, 3 };
int []multipliers = { 3, 2, 1 };
System.out.println(getMaxScore(array, multipliers));
}
}
// This code is contributed by AnkThon
Python3
# Python program for the above approach # Function to find the maximum score # recursively def getMaxScore(array, multipliers): # Depth first search def dfs(start, end, index): if index == len(multipliers): return 0 # Pick left left = multipliers[index] * array[start] + \ dfs(start + 1, end, index + 1) # Pick right right = multipliers[index] * array[end] + \ dfs(start, end - 1, index + 1) return max(right, left) return dfs(0, len(array) - 1, 0) # Driver Code if __name__ == "__main__": array = [1, 2, 3] multipliers = [3, 2, 1] print(getMaxScore(array, multipliers))
C#
// C# program for the above approach
using System;
public class GFG
{
// Function to find the maximum score
// using dynamic programming and
// memoization
static int getMaxScore(int[] array, int[] multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.Length;
int N = array.Length;
int remain = N - M;
int[] dp = new int[M + 1];
for (int i = 0; i < M; ++i)
{
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j)
{
dp[j] = Math.Max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
public static void Main(String[] args)
{
int[] array = { 1, 2, 3 };
int[] multipliers = { 3, 2, 1 };
Console.Write(getMaxScore(array, multipliers));
}
}
// This code is contributed by gfgking.
Javascript
<script>
// Javascript program for the above approach
// Function to find the maximum score
// using dynamic programming and
// memoization
function getMaxScore(array, multipliers) {
// M is the number of elements needed to pick
let M = multipliers.length, N = array.length;
let remain = N - M;
let dp = new Array(M + 1).fill(0);
for (let i = 0; i < M; ++i) {
let mm = multipliers[M - i - 1];
for (let j = 0; j < M - i; ++j) {
dp[j] = Math.max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
let array = [1, 2, 3];
let multipliers = [3, 2, 1];
document.write(getMaxScore(array, multipliers));
// This code is contributed by gfgking
</script>
14
Tiempo Complejidad: O(2 M )
Espacio Auxiliar: O(1)
Enfoque eficiente: la solución se basa en la programación dinámica, ya que contiene ambas propiedades: subestructura óptima y subproblemas superpuestos . Suponga que dp[i][j] es el resultado máximo actual que se puede obtener de un subarreglo, donde i es el índice inicial y j es el final. En cualquier etapa, hay dos opciones:
elige el primero: dp[i + 1][j] + curr_weight * array[i]
elige el último: dp[i][j – 1] + curr_weight * array[j]
El resultado será el máximo de ambos. Siga los pasos a continuación para resolver el problema utilizando la búsqueda y memorización en profundidad :
- Inicializar una variable permanece como NM .
- Inicialice una array dp[] de tamaño M+1 con valores 0.
- Iterar sobre el rango [0, M) usando la variable i y realizar los siguientes pasos:
- Inicializa una variable mm como multiplicadores[-i-1] .
- Iterar sobre el rango [0, Mi) usando la variable j y realizar los siguientes pasos:
- Establezca el valor de dp[j] como el máximo de mm*array[j] + dp[j+1] o mm*array[j+remain] + dp[j].
- Aumenta el valor de permanecer en 1.
- Después de realizar los pasos anteriores, imprima el valor de dp[0] como respuesta.
A continuación se muestra la implementación del enfoque anterior:
C++
// Java program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum score
// using dynamic programming and
// memoization
int getMaxScore(vector<int>& array,
vector<int>& multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.size(), N = array.size();
int remain = N - M;
vector<int> dp(M + 1, 0);
for (int i = 0; i < M; ++i) {
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j) {
dp[j] = max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
int main ()
{
vector<int> array = { 1, 2, 3 };
vector<int> multipliers = { 3, 2, 1 };
cout << getMaxScore(array, multipliers);
}
// This code is contributed by shikhasingrajput
Java
// Java program for the above approach
import java.util.*;
public class GFG {
// Function to find the maximum score
// using dynamic programming and
// memoization
static int getMaxScore(int []array,int []multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.length;
int N = array.length;
int remain = N - M;
int dp[] = new int[M + 1];
for (int i = 0; i < M; ++i) {
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j) {
dp[j] = Math.max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
public static void main (String[] args)
{
int []array = { 1, 2, 3 };
int []multipliers = { 3, 2, 1 };
System.out.println(getMaxScore(array, multipliers));
}
}
// This code is contributed by Samim Hossain Mondal.
Python3
# Python program for the above approach # Function to find the maximum score # using dynamic programming and # memoization def getMaxScore(array, multipliers): # M is the number of elements needed to pick M, N = len(multipliers), len(array) remain = N - M dp = [0] * (M + 1) for i in range(M): mm = multipliers[-i - 1] for j in range(M - i): dp[j] = max(mm * array[j] + dp[j + 1], mm * array[j + remain] + dp[j]) remain += 1 return dp[0] # Driver Code if __name__ == "__main__": array = [1, 2, 3] multipliers = [3, 2, 1] print(getMaxScore(array, multipliers))
C#
// C# program for the above approach
using System;
public class GFG {
// Function to find the maximum score
// using dynamic programming and
// memoization
static int getMaxScore(int[] array, int[] multipliers)
{
// M is the number of elements needed to pick
int M = multipliers.Length;
int N = array.Length;
int remain = N - M;
int[] dp = new int[M + 1];
for (int i = 0; i < M; ++i) {
int mm = multipliers[M - i - 1];
for (int j = 0; j < M - i; ++j) {
dp[j] = Math.Max(mm * array[j] + dp[j + 1],
mm * array[j + remain]
+ dp[j]);
}
remain += 1;
}
return dp[0];
}
// Driver Code
public static void Main(string[] args)
{
int[] array = { 1, 2, 3 };
int[] multipliers = { 3, 2, 1 };
Console.WriteLine(getMaxScore(array, multipliers));
}
}
// This code is contributed by ukasp.
Javascript
<script>
// JavaScript Program to implement
// the above approach
//Function to find the maximum score
//using dynamic programming and
//memoization
function getMaxScore(array, multipliers) {
//M is the number of elements needed to pick
M = multipliers.length
N = array.length
remain = N - M
dp = new Array(M + 1).fill(0)
for (let i = 0; i < M; i++) {
mm = multipliers[M - i - 1]
for (j = 0; j < M - i; j++)
dp[j] = Math.max(mm * array[j] + dp[j + 1],
mm * array[j + remain] + dp[j])
remain += 1
}
return dp[0]
}
array = [1, 2, 3]
multipliers = [3, 2, 1]
document.write(getMaxScore(array, multipliers))
// This code is contributed by Potta Lokesh
</script>
14
Complejidad temporal: O(M*M)
Espacio auxiliar: O(M)
Publicación traducida automáticamente
Artículo escrito por maheswaripiyush9 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA