Se le da una bolsa de tamaño W kg y se le proporcionan costos de paquetes de diferentes pesos de naranjas en la array costo[] donde costo[i] es básicamente el costo de ‘i’ kg paquete de naranjas. Donde costo[i] = -1 significa que el paquete de ‘i’
kg de naranja no está disponible Encuentre el costo total mínimo para comprar exactamente W kg de naranjas y, si no es posible comprar exactamente W kg de naranjas, imprima -1. Se puede suponer que hay un suministro infinito de todos los tipos de paquetes disponibles.
Nota: la array comienza desde el índice 1.
Ejemplos:
Input : W = 5, cost[] = {20, 10, 4, 50, 100}
Output : 14
We can choose two oranges to minimize cost. First
orange of 2Kg and cost 10. Second orange of 3Kg
and cost 4.
Input : W = 5, cost[] = {1, 10, 4, 50, 100}
Output : 5
We can choose five oranges of weight 1 kg.
Input : W = 5, cost[] = {1, 2, 3, 4, 5}
Output : 5
Costs of 1, 2, 3, 4 and 5 kg packets are 1, 2, 3,
4 and 5 Rs respectively.
We choose packet of 5kg having cost 5 for minimum
cost to get 5Kg oranges.
Input : W = 5, cost[] = {-1, -1, 4, 5, -1}
Output : -1
Packets of size 1, 2 and 5 kg are unavailable
because they have cost -1. Cost of 3 kg packet
is 4 Rs and of 4 kg is 5 Rs. Here we have only
weights 3 and 4 so by using these two we can
not make exactly W kg weight, therefore answer
is -1.
Este problema se puede reducir a Ilimitado Knapsac k. Entonces, en la array de costos, primero ignoramos aquellos paquetes que no están disponibles, es decir; cost es -1 y luego recorrer la array de costos y crear dos arrays val[] para almacenar el costo de ‘i’ kg paquete de naranja y wt[] para almacenar el peso del paquete correspondiente. Supongamos que cost[i] = 50, por lo que el peso del paquete será i y el coste será 50.
Algoritmo:
- Cree la array min_cost[n+1][W+1], donde n es el número de paquetes de naranja ponderados distintos y W es la capacidad máxima de la bolsa.
- Inicialice la fila 0 con INF (infinito) y la columna 0 con 0.
- Ahora llena la array
- si wt[i-1] > j entonces min_cost[i][j] = min_cost[i-1][j] ;
- si wt[i-1] <= j entonces min_cost[i][j] = min(min_cost[i-1][j], val[i-1] + min_cost[i][j-wt[i-1 ]]);
- Si min_cost[n][W]==INF, la salida será -1 porque esto significa que no podemos hacer que el peso sea W usando estos pesos; de lo contrario, la salida será min_cost[n][W] .
C++
// C++ program to find minimum cost to get exactly
// W Kg with given packets
#include<bits/stdc++.h>
#define INF 1000000
using namespace std;
// cost[] initial cost array including unavailable packet
// W capacity of bag
int MinimumCost(int cost[], int n, int W)
{
// val[] and wt[] arrays
// val[] array to store cost of 'i' kg packet of orange
// wt[] array weight of packet of orange
vector<int> val, wt;
// traverse the original cost[] array and skip
// unavailable packets and make val[] and wt[]
// array. size variable tells the available number
// of distinct weighted packets
int size = 0;
for (int i=0; i<n; i++)
{
if (cost[i]!= -1)
{
val.push_back(cost[i]);
wt.push_back(i+1);
size++;
}
}
n = size;
int min_cost[n+1][W+1];
// fill 0th row with infinity
for (int i=0; i<=W; i++)
min_cost[0][i] = INF;
// fill 0'th column with 0
for (int i=1; i<=n; i++)
min_cost[i][0] = 0;
// now check for each weight one by one and fill the
// matrix according to the condition
for (int i=1; i<=n; i++)
{
for (int j=1; j<=W; j++)
{
// wt[i-1]>j means capacity of bag is
// less than weight of item
if (wt[i-1] > j)
min_cost[i][j] = min_cost[i-1][j];
// here we check we get minimum cost either
// by including it or excluding it
else
min_cost[i][j] = min(min_cost[i-1][j],
min_cost[i][j-wt[i-1]] + val[i-1]);
}
}
// exactly weight W can not be made by given weights
return (min_cost[n][W]==INF)? -1: min_cost[n][W];
}
// Driver program to run the test case
int main()
{
int cost[] = {1, 2, 3, 4, 5}, W = 5;
int n = sizeof(cost)/sizeof(cost[0]);
cout << MinimumCost(cost, n, W);
return 0;
}
Java
// Java Code for Minimum cost to
// fill given weight in a bag
import java.util.*;
class GFG {
// cost[] initial cost array including
// unavailable packet W capacity of bag
public static int MinimumCost(int cost[], int n,
int W)
{
// val[] and wt[] arrays
// val[] array to store cost of 'i' kg
// packet of orange wt[] array weight of
// packet of orange
Vector<Integer> val = new Vector<Integer>();
Vector<Integer> wt = new Vector<Integer>();
// traverse the original cost[] array and skip
// unavailable packets and make val[] and wt[]
// array. size variable tells the available
// number of distinct weighted packets
int size = 0;
for (int i = 0; i < n; i++)
{
if (cost[i] != -1)
{
val.add(cost[i]);
wt.add(i + 1);
size++;
}
}
n = size;
int min_cost[][] = new int[n+1][W+1];
// fill 0th row with infinity
for (int i = 0; i <= W; i++)
min_cost[0][i] = Integer.MAX_VALUE;
// fill 0'th column with 0
for (int i = 1; i <= n; i++)
min_cost[i][0] = 0;
// now check for each weight one by one and
// fill the matrix according to the condition
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= W; j++)
{
// wt[i-1]>j means capacity of bag is
// less than weight of item
if (wt.get(i-1) > j)
min_cost[i][j] = min_cost[i-1][j];
// here we check we get minimum cost
// either by including it or excluding
// it
else
min_cost[i][j] = Math.min(min_cost[i-1][j],
min_cost[i][j-wt.get(i-1)] +
val.get(i-1));
}
}
// exactly weight W can not be made by
// given weights
return (min_cost[n][W] == Integer.MAX_VALUE)? -1:
min_cost[n][W];
}
/* Driver program to test above function */
public static void main(String[] args)
{
int cost[] = {1, 2, 3, 4, 5}, W = 5;
int n = cost.length;
System.out.println(MinimumCost(cost, n, W));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python program to find minimum cost to get exactly # W Kg with given packets INF = 1000000 # cost[] initial cost array including unavailable packet # W capacity of bag def MinimumCost(cost, n, W): # val[] and wt[] arrays # val[] array to store cost of 'i' kg packet of orange # wt[] array weight of packet of orange val = list() wt= list() # traverse the original cost[] array and skip # unavailable packets and make val[] and wt[] # array. size variable tells the available number # of distinct weighted packets. size = 0 for i in range(n): if (cost[i] != -1): val.append(cost[i]) wt.append(i+1) size += 1 n = size min_cost = [[0 for i in range(W+1)] for j in range(n+1)] # fill 0th row with infinity for i in range(W+1): min_cost[0][i] = INF # fill 0th column with 0 for i in range(1, n+1): min_cost[i][0] = 0 # now check for each weight one by one and fill the # matrix according to the condition for i in range(1, n+1): for j in range(1, W+1): # wt[i-1]>j means capacity of bag is # less than weight of item if (wt[i-1] > j): min_cost[i][j] = min_cost[i-1][j] # here we check we get minimum cost either # by including it or excluding it else: min_cost[i][j] = min(min_cost[i-1][j], min_cost[i][j-wt[i-1]] + val[i-1]) # exactly weight W can not be made by given weights if(min_cost[n][W] == INF): return -1 else: return min_cost[n][W] # Driver program to run the test case cost = [1, 2, 3, 4, 5] W = 5 n = len(cost) print(MinimumCost(cost, n, W)) # This code is contributed by Soumen Ghosh.
C#
// C# Code for Minimum cost to
// fill given weight in a bag
using System;
using System.Collections.Generic;
class GFG {
// cost[] initial cost array including
// unavailable packet W capacity of bag
public static int MinimumCost(int []cost, int n,
int W)
{
// val[] and wt[] arrays
// val[] array to store cost of 'i' kg
// packet of orange wt[] array weight of
// packet of orange
List<int> val = new List<int>();
List<int> wt = new List<int>();
// traverse the original cost[] array and skip
// unavailable packets and make val[] and wt[]
// array. size variable tells the available
// number of distinct weighted packets
int size = 0;
for (int i = 0; i < n; i++)
{
if (cost[i] != -1)
{
val.Add(cost[i]);
wt.Add(i + 1);
size++;
}
}
n = size;
int [,]min_cost = new int[n+1,W+1];
// fill 0th row with infinity
for (int i = 0; i <= W; i++)
min_cost[0,i] = int.MaxValue;
// fill 0'th column with 0
for (int i = 1; i <= n; i++)
min_cost[i,0] = 0;
// now check for each weight one by one and
// fill the matrix according to the condition
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= W; j++)
{
// wt[i-1]>j means capacity of bag is
// less than weight of item
if (wt[i-1] > j)
min_cost[i,j] = min_cost[i-1,j];
// here we check we get minimum cost
// either by including it or excluding
// it
else
min_cost[i,j] = Math.Min(min_cost[i-1,j],
min_cost[i,j-wt[i-1]] + val[i-1]);
}
}
// exactly weight W can not be made by
// given weights
return (min_cost[n,W] == int.MaxValue )? -1: min_cost[n,W];
}
/* Driver program to test above function */
public static void Main()
{
int []cost = {1, 2, 3, 4, 5};
int W = 5;
int n = cost.Length;
Console.WriteLine(MinimumCost(cost, n, W));
}
}
// This code is contributed by Ryuga
PHP
<?php
// PHP program to find minimum cost to
// get exactly W Kg with given packets
$INF = 1000000;
// cost[] initial cost array including
// unavailable packet W capacity of bag
function MinimumCost(&$cost, $n, $W)
{
global $INF;
// val[] and wt[] arrays
// val[] array to store cost of 'i'
// kg packet of orange
// wt[] array weight of packet of orange
$val = array();
$wt = array();
// traverse the original cost[] array
// and skip unavailable packets and
// make val[] and wt[] array. size
// variable tells the available number
// of distinct weighted packets
$size = 0;
for ($i = 0; $i < $n; $i++)
{
if ($cost[$i] != -1)
{
array_push($val, $cost[$i]);
array_push($wt, $i + 1);
$size++;
}
}
$n = $size;
$min_cost = array_fill(0, $n + 1,
array_fill(0, $W + 1, NULL));
// fill 0th row with infinity
for ($i = 0; $i <= $W; $i++)
$min_cost[0][$i] = $INF;
// fill 0'th column with 0
for ($i = 1; $i <= $n; $i++)
$min_cost[$i][0] = 0;
// now check for each weight one by
// one and fill the matrix according
// to the condition
for ($i = 1; $i <= $n; $i++)
{
for ($j = 1; $j <= $W; $j++)
{
// wt[i-1]>j means capacity of bag
// is less than weight of item
if ($wt[$i - 1] > $j)
$min_cost[$i][$j] = $min_cost[$i - 1][$j];
// here we check we get minimum
// cost either by including it
// or excluding it
else
$min_cost[$i][$j] = min($min_cost[$i - 1][$j],
$min_cost[$i][$j - $wt[$i - 1]] +
$val[$i - 1]);
}
}
// exactly weight W can not be made
// by given weights
if ($min_cost[$n][$W] == $INF)
return -1;
else
return $min_cost[$n][$W];
}
// Driver Code
$cost = array(1, 2, 3, 4, 5);
$W = 5;
$n = sizeof($cost);
echo MinimumCost($cost, $n, $W);
// This code is contributed by ita_c
?>
Javascript
<script>
// Javascript program to find minimum cost to get exactly
// W Kg with given packets
let INF = 1000000;
// cost[] initial cost array including unavailable packet
// W capacity of bag
function MinimumCost(cost, n, W)
{
// val[] and wt[] arrays
// val[] array to store cost of 'i' kg packet of orange
// wt[] array weight of packet of orange
let val = [], wt = [];
// traverse the original cost[] array and skip
// unavailable packets and make val[] and wt[]
// array. size variable tells the available number
// of distinct weighted packets
let size = 0;
for (let i=0; i<n; i++)
{
if (cost[i]!= -1)
{
val.push(cost[i]);
wt.push(i+1);
size++;
}
}
n = size;
let min_cost = new Array(n+1);
for(let i = 0; i < n + 1; i++)
{
min_cost[i] = new Array(W + 1);
}
// fill 0th row with infinity
for (let i=0; i<=W; i++)
min_cost[0][i] = INF;
// fill 0'th column with 0
for (let i=1; i<=n; i++)
min_cost[i][0] = 0;
// now check for each weight one by one and fill the
// matrix according to the condition
for (let i=1; i<=n; i++)
{
for (let j=1; j<=W; j++)
{
// wt[i-1]>j means capacity of bag is
// less than weight of item
if (wt[i-1] > j)
min_cost[i][j] = min_cost[i-1][j];
// here we check we get minimum cost either
// by including it or excluding it
else
min_cost[i][j] = Math.min(min_cost[i-1][j],
min_cost[i][j-wt[i-1]] + val[i-1]);
}
}
// exactly weight W can not be made by given weights
return (min_cost[n][W]==INF)? -1: min_cost[n][W];
}
// Driver code
let cost = [1, 2, 3, 4, 5], W = 5;
let n = cost.length;
document.write(MinimumCost(cost, n, W));
// This code is contributed by suresh07.
</script>
Producción
5
Solución de espacio optimizado Si observamos más de cerca este problema, podemos notar que se trata de una variación del Problema de corte de varillas . En lugar de hacer la maximización, aquí tenemos que hacer la minimización.
C++
// C++ program to find minimum cost to
// get exactly W Kg with given packets
#include<bits/stdc++.h>
using namespace std;
/* Returns the best obtainable price for
a rod of length n and price[] as prices
of different pieces */
int minCost(int cost[], int n)
{
int dp[n+1];
dp[0] = 0;
// Build the table val[] in bottom up
// manner and return the last entry
// from the table
for (int i = 1; i<=n; i++)
{
int min_cost = INT_MAX;
for (int j = 0; j < i; j++)
if(cost[j]!=-1)
min_cost = min(min_cost, cost[j] + dp[i-j-1]);
dp[i] = min_cost;
}
return dp[n];
}
/* Driver code */
int main()
{
int cost[] = {20, 10, 4, 50, 100};
int W = sizeof(cost)/sizeof(cost[0]);
cout << minCost(cost, W);
return 0;
}
Java
// Java program to find minimum cost to
// get exactly W Kg with given packets
import java.util.*;
class Main
{
/* Returns the best obtainable price for
a rod of length n and price[] as prices
of different pieces */
public static int minCost(int cost[], int n)
{
int dp[] = new int[n + 1];
dp[0] = 0;
// Build the table val[] in bottom up
// manner and return the last entry
// from the table
for (int i = 1; i <= n; i++)
{
int min_cost = Integer.MAX_VALUE;
for (int j = 0; j < i; j++)
if(cost[j]!=-1) {
min_cost = Math.min(min_cost, cost[j] + dp[i - j - 1]);
}
dp[i] = min_cost;
}
return dp[n];
}
public static void main(String[] args) {
int cost[] = {10,-1,-1,-1,-1};
int W = cost.length;
System.out.print(minCost(cost, W));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program to find minimum cost to # get exactly W Kg with given packets import sys # Returns the best obtainable price for # a rod of length n and price[] as prices # of different pieces def minCost(cost, n): dp = [0 for i in range(n + 1)] # Build the table val[] in bottom up # manner and return the last entry # from the table for i in range(1, n + 1): min_cost = sys.maxsize for j in range(i): if cost[j]!=-1: min_cost = min(min_cost, cost[j] + dp[i - j - 1]) dp[i] = min_cost return dp[n] # Driver code cost = [ 10,-1,-1,-1,-1 ] W = len(cost) print(minCost(cost, W)) # This code is contributed by rag2127
C#
// C# program to find minimum cost to
// get exactly W Kg with given packets
using System;
class GFG {
/* Returns the best obtainable price for
a rod of length n and price[] as prices
of different pieces */
static int minCost(int[] cost, int n)
{
int[] dp = new int[n + 1];
dp[0] = 0;
// Build the table val[] in bottom up
// manner and return the last entry
// from the table
for (int i = 1; i <= n; i++)
{
int min_cost = Int32.MaxValue;
for (int j = 0; j < i; j++)
if(cost[j]!=-1)
min_cost = Math.Min(min_cost,
cost[j] + dp[i - j - 1]);
dp[i] = min_cost;
}
return dp[n];
}
// Driver code
static void Main() {
int[] cost = {20, 10, 4, 50, 100};
int W = cost.Length;
Console.Write(minCost(cost, W));
}
}
// This code is contributed by divyesh072019
Javascript
<script>
// Javascript program to find minimum cost to
// get exactly W Kg with given packets
/* Returns the best obtainable price for
a rod of length n and price[] as prices
of different pieces */
function minCost(cost, n)
{
let dp = new Array(n+1);
dp[0] = 0;
// Build the table val[] in bottom up
// manner and return the last entry
// from the table
for (let i = 1; i<=n; i++)
{
let min_cost = Number.MAX_VALUE;
for (let j = 0; j < i; j++)
if(j < n)
min_cost = Math.min(min_cost, cost[j] + dp[i-j-1]);
dp[i] = min_cost;
}
return dp[n];
}
let cost = [20, 10, 4, 50, 100];
let W = cost.length;
document.write(minCost(cost, W));
</script>
Producción
14
Enfoque de arriba hacia abajo: también podemos resolver el problema utilizando la memorización.
C++
// C++ program to find minimum cost to
// get exactly W Kg with given packets
#include <bits/stdc++.h>
using namespace std;
int helper(vector<int>& cost, vector<int>& weight, int n,
int w, vector<vector<int> >& dp)
{
// base cases
if (w == 0)
return 0;
if (w < 0 or n <= 0)
return INT_MAX;
if (dp[n][w] != -1)
return dp[n][w];
if (cost[n - 1] < 0)
return dp[n][w]
= min(INT_MAX,
helper(cost, weight, n - 1, w, dp));
if (weight[n - 1] <= w) {
return dp[n][w]
= min(cost[n - 1]
+ helper(cost, weight, n,
w - weight[n - 1], dp),
helper(cost, weight, n - 1, w, dp));
}
return dp[n][w] = helper(cost, weight, n - 1, w, dp);
}
int minCost(vector<int>& cost, int W)
{
int N = cost.size();
// Your code goes here
vector<int> weight(N);
// create the weight array
for (int i = 1; i <= N; i++) {
weight[i - 1] = i;
}
// initialize the dp array
vector<vector<int> > dp(N + 1, vector<int>(W + 1, -1));
int res = helper(cost, weight, N, W, dp);
// return -1 if result is MAX
return (res == INT_MAX) ? -1 : res;
}
/* Driver code */
int main()
{
vector<int> cost = { 20, 10, 4, 50, 100 };
int W = cost.size();
cout << minCost(cost, W);
return 0;
}
Java
// Java program to find minimum cost to
// get exactly W Kg with given packets
import java.io.*;
class GFG {
public static int helper(int cost[], int weight[],
int n, int w, int dp[][])
{
// base cases
if (w == 0)
return 0;
if (w < 0 || n <= 0)
return Integer.MAX_VALUE;
if (dp[n][w] != -1)
return dp[n][w];
if (cost[n - 1] < 0)
return dp[n][w] = Math.min(
Integer.MAX_VALUE,
helper(cost, weight, n - 1, w, dp));
if (weight[n - 1] <= w) {
return dp[n][w] = Math.min(
cost[n - 1]
+ helper(cost, weight, n,
w - weight[n - 1], dp),
helper(cost, weight, n - 1, w, dp));
}
return dp[n][w]
= helper(cost, weight, n - 1, w, dp);
}
public static int minCost(int cost[], int W)
{
int N = cost.length;
int weight[] = new int[N];
// create the weight array
for (int i = 1; i <= N; i++) {
weight[i - 1] = i;
}
// initialize the dp array
int dp[][] = new int[N + 1][W + 1];
for (int i = 0; i < N + 1; i++)
for (int j = 0; j < W + 1; j++)
dp[i][j] = -1;
int res = helper(cost, weight, N, W, dp);
// return -1 if result is MAX
return (res == Integer.MAX_VALUE) ? -1 : res;
}
// Driver Code
public static void main(String[] args)
{
int cost[] = { 20, 10, 4, 50, 100 };
int W = cost.length;
System.out.print(minCost(cost, W));
}
}
// This code is contributed by Rohit Pradhan
Python3
# Python3 program to find minimum cost to # get exactly W Kg with given packets import sys def helper(cost, weight, n, w, dp): # base cases if (w == 0): return 0 if (w < 0 or n <= 0): return sys.maxsize if (dp[n][w] != -1): return dp[n][w] if (cost[n - 1] < 0): dp[n][w] = min(sys.maxsize, helper(cost, weight, n - 1, w, dp)) return dp[n][w] if (weight[n - 1] <= w): dp[n][w] = min(cost[n - 1] + helper(cost, weight, n, w - weight[n - 1], dp), helper(cost, weight, n - 1, w, dp)) return dp[n][w] dp[n][w] = helper(cost, weight, n - 1, w, dp) return dp[n][w] def minCost(cost, W): N = len(cost) weight = [0 for i in range(N)] # create the weight array for i in range(1,N + 1): weight[i - 1] = i # initialize the dp array dp = [[-1 for i in range(W + 1)]for j in range(N + 1)] res = helper(cost, weight, N, W, dp) # return -1 if result is MAX return -1 if(res == sys.maxsize) else res # Driver code cost = [ 20, 10, 4, 50, 100 ] W = len(cost) print(minCost(cost, W)) # This code is contributed by shinjanpatra
C#
// C# program to find minimum cost to
// get exactly W Kg with given packets
using System;
class GFG
{
static int helper(int[] cost, int[] weight,
int n, int w, int[,] dp)
{
// base cases
if (w == 0)
return 0;
if (w < 0 || n <= 0)
return Int32.MaxValue;
if (dp[n,w] != -1)
return dp[n,w];
if (cost[n - 1] < 0)
return dp[n,w] = Math.Min(
Int32.MaxValue,
helper(cost, weight, n - 1, w, dp));
if (weight[n - 1] <= w)
{
return dp[n,w] = Math.Min(
cost[n - 1]
+ helper(cost, weight, n,
w - weight[n - 1], dp),
helper(cost, weight, n - 1, w, dp));
}
return dp[n,w]
= helper(cost, weight, n - 1, w, dp);
}
static int minCost(int[] cost, int W)
{
int N = cost.Length;
int[] weight = new int[N];
// create the weight array
for (int i = 1; i <= N; i++)
{
weight[i - 1] = i;
}
// initialize the dp array
int[,] dp = new int[N + 1, W + 1];
for (int i = 0; i < N + 1; i++)
for (int j = 0; j < W + 1; j++)
dp[i,j] = -1;
int res = helper(cost, weight, N, W, dp);
// return -1 if result is MAX
return (res == Int32.MaxValue) ? -1 : res;
}
// Driver Code
static public void Main()
{
int[] cost = { 20, 10, 4, 50, 100 };
int W = cost.Length;
Console.Write(minCost(cost, W));
}
}
// This code is contributed by kothavvsaakash
Javascript
<script>
// JavaScript program to find minimum cost to
// get exactly W Kg with given packets
function helper(cost, weight, n, w, dp)
{
// base cases
if (w == 0)
return 0;
if (w < 0 || n <= 0)
return Number.MAX_VALUE;
if (dp[n][w] != -1)
return dp[n][w];
if (cost[n - 1] < 0)
return dp[n][w]
= Math.min(Number.MAX_VALUE,
helper(cost, weight, n - 1, w, dp));
if (weight[n - 1] <= w) {
return dp[n][w]
= Math.min(cost[n - 1]
+ helper(cost, weight, n,
w - weight[n - 1], dp),
helper(cost, weight, n - 1, w, dp));
}
return dp[n][w] = helper(cost, weight, n - 1, w, dp);
}
function minCost(cost,W)
{
let N = cost.length;
// Your code goes here
let weight = new Array(N);
// create the weight array
for (let i = 1; i <= N; i++) {
weight[i - 1] = i;
}
// initialize the dp array
let dp = new Array(N + 1).fill(-1).map(()=>new Array(W + 1).fill(-1));
let res = helper(cost, weight, N, W, dp);
// return -1 if result is MAX
return (res == Number.MAX_VALUE) ? -1 : res;
}
/* Driver code */
let cost = [ 20, 10, 4, 50, 100 ];
let W = cost.length;
document.write(minCost(cost, W),"</br>");
// This code is contributed by shinjanpatra
</script>
Producción
14
Este artículo es una contribución de Shashank Mishra (Gullu) . Este artículo es revisado por el equipo GeeksForGeeks.
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