Dado un valor N, si queremos dar cambio por N centavos, y tenemos un suministro infinito de cada una de las monedas valoradas en S = { S1, S2, .. , Sm}, ¿de cuántas formas podemos hacer el cambio? El orden de las monedas no importa.
Por ejemplo, para N = 4 y S = {1,2,3}, hay cuatro soluciones: {1,1,1,1},{1,1,2},{2,2},{1, 3}. Entonces, la salida debería ser 4. Para N = 10 y S = {2, 5, 3, 6}, hay cinco soluciones: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} y {5,5}. Entonces la salida debería ser 5.
1) Subestructura óptima
Para contar el número total de soluciones, podemos dividir todas las soluciones del conjunto en dos conjuntos.
1) Soluciones que no contienen moneda mth (o Sm).
2) Soluciones que contengan al menos un Sm.
Sea count(S[], m, n) la función para contar el número de soluciones, entonces se puede escribir como la suma de count(S[], m-1, n) y count(S[], m, n-Sm).
Por lo tanto, el problema tiene una propiedad de subestructura óptima ya que el problema se puede resolver usando soluciones a los subproblemas.
2) Subproblemas superpuestos
A continuación se muestra una implementación recursiva simple del problema de cambio de moneda. La implementación simplemente sigue la estructura recursiva mencionada anteriormente.
3) Enfoque (Algoritmo)
Mira, aquí cada moneda de una denominación dada puede venir un número infinito de veces. (Se permite la repetición), esto es lo que llamamos MOCHILA SIN LÍMITES. Tenemos 2 opciones para una moneda de una denominación particular, ya sea i) para incluir, o ii) para excluir. Pero aquí, el proceso de inclusión no es de una sola vez; podemos incluir cualquier denominación cualquier número de veces hasta N<0.
Básicamente, si estamos en s[m-1], podemos tomar tantas instancias de esa moneda (inclusión ilimitada), es decir, count(S, m, n – S[m-1]) ; luego pasamos a s[m-2]. Después de pasar a s[m-2], no podemos retroceder y no podemos tomar decisiones para s[m-1], es decir, count(S, m-1, n) .
Finalmente, como tenemos que encontrar el número total de formas, sumaremos estas 2 opciones posibles, es decir, count(S, m, n – S[m-1] ) + count(S, m-1, n ) ; que será nuestra respuesta requerida.
C++
// Recursive C++ program for
// coin change problem.
#include <bits/stdc++.h>
using namespace std;
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
int count(int S[], int m, int n)
{
// If n is 0 then there is 1 solution
// (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if (n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if (m <= 0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return count(S, m - 1, n) +
count(S, m, n - S[m - 1]);
}
// Driver code
int main()
{
int i, j;
int arr[] = { 1, 2, 3 };
int m = sizeof(arr) / sizeof(arr[0]);
cout << " " << count(arr, m, 4);
return 0;
}
// This code is contributed by shivanisinghss2110
C
// Recursive C program for
// coin change problem.
#include<stdio.h>
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
int count( int S[], int m, int n )
{
// If n is 0 then there is 1 solution
// (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if (n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if (m <=0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}
// Driver program to test above function
int main()
{
int i, j;
int arr[] = {1, 2, 3};
int m = sizeof(arr)/sizeof(arr[0]);
printf("%d ", count(arr, m, 4));
getchar();
return 0;
}
Java
// Recursive JAVA program for
// coin change problem.
import java.util.*;
class GFG
{
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
static int count(int S[], int m, int n)
{
// If n is 0 then there is 1 solution
// (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if (n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if (m <= 0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return count(S, m - 1, n) +
count(S, m, n - S[m - 1]);
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 2, 3 };
int m = arr.length;
System.out.println(count(arr, m, 4));
}
}
// This code is contributed by jyoti369
Python3
# Recursive Python3 program for # coin change problem. # Returns the count of ways we can sum # S[0...m-1] coins to get sum n def count(S, m, n ): # If n is 0 then there is 1 # solution (do not include any coin) if (n == 0): return 1 # If n is less than 0 then no # solution exists if (n < 0): return 0; # If there are no coins and n # is greater than 0, then no # solution exist if (m <=0): return 0 # count is sum of solutions (i) # including S[m-1] (ii) excluding S[m-1] return count( S, m - 1, n ) + count( S, m, n-S[m-1] ); # Driver program to test above function arr = [1, 2, 3] m = len(arr) print(count(arr, m, 4)) # This code is contributed by Smitha Dinesh Semwal
C#
// Recursive C# program for
// coin change problem.
using System;
class GFG
{
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
static int count( int []S, int m, int n )
{
// If n is 0 then there is 1 solution
// (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if (n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if (m <=0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return count( S, m - 1, n ) +
count( S, m, n - S[m - 1] );
}
// Driver program
public static void Main()
{
int []arr = {1, 2, 3};
int m = arr.Length;
Console.Write( count(arr, m, 4));
}
}
// This code is contributed by Sam007
PHP
<?php
// Recursive PHP program for
// coin change problem.
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
function coun($S, $m, $n)
{
// If n is 0 then there is
// 1 solution (do not include
// any coin)
if ($n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if ($n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if ($m <= 0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return coun($S, $m - 1,$n ) +
coun($S, $m, $n - $S[$m - 1] );
}
// Driver Code
$arr = array(1, 2, 3);
$m = count($arr);
echo coun($arr, $m, 4);
// This code is contributed by Sam007
?>
Javascript
<script>
// Recursive javascript program for
// coin change problem.
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
function count(S , m , n )
{
// If n is 0 then there is 1 solution
// (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no
// solution exists
if (n < 0)
return 0;
// If there are no coins and n
// is greater than 0, then no
// solution exist
if (m <=0)
return 0;
// count is sum of solutions (i)
// including S[m-1] (ii) excluding S[m-1]
return count( S, m - 1, n ) +
count( S, m, n - S[m - 1] );
}
// Driver program to test above function
var arr = [1, 2, 3];
var m = arr.length;
document.write( count(arr, m, 4));
// This code is contributed by Amit Katiyar
</script>
Producción
4
Complejidad del tiempo: O(2 n )
Complejidad espacial: O (objetivo): espacio de pila auxiliar
Cabe señalar que la función anterior calcula los mismos subproblemas una y otra vez. Vea el siguiente árbol de recursión para S = {1, 2, 3} y n = 5.
La función C({1}, 3) se llama dos veces. Si dibujamos el árbol completo, podemos ver que hay muchos subproblemas que se llaman más de una vez.
C() --> count()
C({1,2,3}, 5)
/ \
/ \
C({1,2,3}, 2) C({1,2}, 5)
/ \ / \
/ \ / \
C({1,2,3}, -1) C({1,2}, 2) C({1,2}, 3) C({1}, 5)
/ \ / \ / \
/ \ / \ / \
C({1,2},0) C({1},2) C({1,2},1) C({1},3) C({1}, 4) C({}, 5)
/ \ / \ /\ / \
/ \ / \ / \ / \
. . . . . . C({1}, 3) C({}, 4)
/ \
/ \
. .
Dado que los mismos subproblemas se vuelven a llamar, este problema tiene la propiedad Superposición de subproblemas. Entonces, el problema de cambio de monedas tiene las dos propiedades (ver this y this ) de un problema de programación dinámica. Al igual que otros problemas típicos de programación dinámica (DP) , los cálculos de los mismos subproblemas se pueden evitar mediante la construcción de una tabla de array temporal [] [] de forma ascendente.
Solución de programación dinámica
C++
// C++ program for coin change problem.
#include<bits/stdc++.h>
using namespace std;
int count( int S[], int m, int n )
{
int i, j, x, y;
// We need n+1 rows as the table
// is constructed in bottom up
// manner using the base case 0
// value case (n = 0)
int table[n + 1][m];
// Fill the entries for 0
// value case (n = 0)
for (i = 0; i < m; i++)
table[0][i] = 1;
// Fill rest of the table entries
// in bottom up manner
for (i = 1; i < n + 1; i++)
{
for (j = 0; j < m; j++)
{
// Count of solutions including S[j]
x = (i-S[j] >= 0) ? table[i - S[j]][j] : 0;
// Count of solutions excluding S[j]
y = (j >= 1) ? table[i][j - 1] : 0;
// total count
table[i][j] = x + y;
}
}
return table[n][m - 1];
}
// Driver Code
int main()
{
int arr[] = {1, 2, 3};
int m = sizeof(arr)/sizeof(arr[0]);
int n = 4;
cout << count(arr, m, n);
return 0;
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
C
// C program for coin change problem.
#include<stdio.h>
int count( int S[], int m, int n )
{
int i, j, x, y;
// We need n+1 rows as the table is constructed
// in bottom up manner using the base case 0
// value case (n = 0)
int table[n+1][m];
// Fill the entries for 0 value case (n = 0)
for (i=0; i<m; i++)
table[0][i] = 1;
// Fill rest of the table entries in bottom
// up manner
for (i = 1; i < n+1; i++)
{
for (j = 0; j < m; j++)
{
// Count of solutions including S[j]
x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;
// Count of solutions excluding S[j]
y = (j >= 1)? table[i][j-1]: 0;
// total count
table[i][j] = x + y;
}
}
return table[n][m-1];
}
// Driver program to test above function
int main()
{
int arr[] = {1, 2, 3};
int m = sizeof(arr)/sizeof(arr[0]);
int n = 4;
printf(" %d ", count(arr, m, n));
return 0;
}
Java
/* Dynamic Programming Java implementation of Coin
Change problem */
import java.util.Arrays;
class CoinChange
{
static long countWays(int S[], int m, int n)
{
//Time complexity of this function: O(mn)
//Space Complexity of this function: O(n)
// table[i] will be storing the number of solutions
// for value i. We need n+1 rows as the table is
// constructed in bottom up manner using the base
// case (n = 0)
long[] table = new long[n+1];
// Initialize all table values as 0
Arrays.fill(table, 0); //O(n)
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and update the table[]
// values after the index greater than or equal to
// the value of the picked coin
for (int i=0; i<m; i++)
for (int j=S[i]; j<=n; j++)
table[j] += table[j-S[i]];
return table[n];
}
// Driver Function to test above function
public static void main(String args[])
{
int arr[] = {1, 2, 3};
int m = arr.length;
int n = 4;
System.out.println(countWays(arr, m, n));
}
}
// This code is contributed by Pankaj Kumar
Python
# Dynamic Programming Python implementation of Coin # Change problem def count(S, m, n): # We need n+1 rows as the table is constructed # in bottom up manner using the base case 0 value # case (n = 0) table = [[0 for x in range(m)] for x in range(n+1)] # Fill the entries for 0 value case (n = 0) for i in range(m): table[0][i] = 1 # Fill rest of the table entries in bottom up manner for i in range(1, n+1): for j in range(m): # Count of solutions including S[j] x = table[i - S[j]][j] if i-S[j] >= 0 else 0 # Count of solutions excluding S[j] y = table[i][j-1] if j >= 1 else 0 # total count table[i][j] = x + y return table[n][m-1] # Driver program to test above function arr = [1, 2, 3] m = len(arr) n = 4 print(count(arr, m, n)) # This code is contributed by Bhavya Jain
C#
/* Dynamic Programming C# implementation of Coin
Change problem */
using System;
class GFG
{
static long countWays(int []S, int m, int n)
{
//Time complexity of this function: O(mn)
//Space Complexity of this function: O(n)
// table[i] will be storing the number of solutions
// for value i. We need n+1 rows as the table is
// constructed in bottom up manner using the base
// case (n = 0)
int[] table = new int[n+1];
// Initialize all table values as 0
for(int i = 0; i < table.Length; i++)
{
table[i] = 0;
}
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and update the table[]
// values after the index greater than or equal to
// the value of the picked coin
for (int i = 0; i < m; i++)
for (int j = S[i]; j <= n; j++)
table[j] += table[j - S[i]];
return table[n];
}
// Driver Function
public static void Main()
{
int []arr = {1, 2, 3};
int m = arr.Length;
int n = 4;
Console.Write(countWays(arr, m, n));
}
}
// This code is contributed by Sam007
PHP
<?php
// PHP program for
// coin change problem.
function count1($S, $m, $n)
{
// We need n+1 rows as
// the table is constructed
// in bottom up manner
// using the base case 0
// value case (n = 0)
$table;
for ($i = 0; $i < $n + 1; $i++)
for ($j = 0; $j < $m; $j++)
$table[$i][$j] = 0;
// Fill the entries for
// 0 value case (n = 0)
for ($i = 0; $i < $m; $i++)
$table[0][$i] = 1;
// Fill rest of the table
// entries in bottom up manner
for ($i = 1; $i < $n + 1; $i++)
{
for ($j = 0; $j < $m; $j++)
{
// Count of solutions
// including S[j]
$x = ($i-$S[$j] >= 0) ?
$table[$i - $S[$j]][$j] : 0;
// Count of solutions
// excluding S[j]
$y = ($j >= 1) ?
$table[$i][$j - 1] : 0;
// total count
$table[$i][$j] = $x + $y;
}
}
return $table[$n][$m-1];
}
// Driver Code
$arr = array(1, 2, 3);
$m = count($arr);
$n = 4;
echo count1($arr, $m, $n);
// This code is contributed by mits
?>
Javascript
<script>
/* Dynamic Programming javascript implementation of Coin
Change problem */
function countWays(S , m , n)
{
//Time complexity of this function: O(mn)
//Space Complexity of this function: O(n)
// table[i] will be storing the number of solutions
// for value i. We need n+1 rows as the table is
// constructed in bottom up manner using the base
// case (n = 0)
// Initialize all table values as 0
//O(n)
var table = Array(n+1).fill(0);
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and update the table
// values after the index greater than or equal to
// the value of the picked coin
for (i=0; i<m; i++)
for (j=S[i]; j<=n; j++)
table[j] += table[j-S[i]];
return table[n];
}
// Driver Function to test above function
var arr = [1, 2, 3];
var m = arr.length;
var n = 4;
document.write(countWays(arr, m, n));
// This code is contributed by 29AjayKumar
</script>
Producción
4
Complejidad de tiempo: O(mn)
La siguiente es una versión simplificada del método 2. El espacio auxiliar requerido aquí es O(n) únicamente.
C++
int count( int S[], int m, int n )
{
// table[i] will be storing the number of solutions for
// value i. We need n+1 rows as the table is constructed
// in bottom up manner using the base case (n = 0)
int table[n+1];
// Initialize all table values as 0
memset(table, 0, sizeof(table));
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and update the table[] values
// after the index greater than or equal to the value of the
// picked coin
for(int i=0; i<m; i++)
for(int j=S[i]; j<=n; j++)
table[j] += table[j-S[i]];
return table[n];
}
Java
public static int count( int S[], int m, int n )
{
// table[i] will be storing the number of solutions for
// value i. We need n+1 rows as the table is constructed
// in bottom up manner using the base case (n = 0)
int table[]=new int[n+1];
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and update the table[] values
// after the index greater than or equal to the value of the
// picked coin
for(int i=0; i<m; i++)
for(int j=S[i]; j<=n; j++)
table[j] += table[j-S[i]];
return table[n];
}
Python
# Dynamic Programming Python implementation of Coin # Change problem def count(S, m, n): # table[i] will be storing the number of solutions for # value i. We need n+1 rows as the table is constructed # in bottom up manner using the base case (n = 0) # Initialize all table values as 0 table = [0 for k in range(n+1)] # Base case (If given value is 0) table[0] = 1 # Pick all coins one by one and update the table[] values # after the index greater than or equal to the value of the # picked coin for i in range(0,m): for j in range(S[i],n+1): table[j] += table[j-S[i]] return table[n] # Driver program to test above function arr = [1, 2, 3] m = len(arr) n = 4 x = count(arr, m, n) print (x) # This code is contributed by Afzal Ansari
C#
// Dynamic Programming C# implementation
// of Coin Change problem
using System;
class GFG
{
static int count(int []S, int m, int n)
{
// table[i] will be storing the
// number of solutions for value i.
// We need n+1 rows as the table
// is constructed in bottom up manner
// using the base case (n = 0)
int [] table = new int[n + 1];
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and
// update the table[] values after
// the index greater than or equal
// to the value of the picked coin
for(int i = 0; i < m; i++)
for(int j = S[i]; j <= n; j++)
table[j] += table[j - S[i]];
return table[n];
}
// Driver Code
public static void Main()
{
int []arr = {1, 2, 3};
int m = arr.Length;
int n = 4;
Console.Write(count(arr, m, n));
}
}
// This code is contributed by Raj
PHP
<?php
function count_1( &$S, $m, $n )
{
// table[i] will be storing the number
// of solutions for value i. We need n+1
// rows as the table is constructed in
// bottom up manner using the base case (n = 0)
$table = array_fill(0, $n + 1, NULl);
// Base case (If given value is 0)
$table[0] = 1;
// Pick all coins one by one and update
// the table[] values after the index
// greater than or equal to the value
// of the picked coin
for($i = 0; $i < $m; $i++)
for($j = $S[$i]; $j <= $n; $j++)
$table[$j] += $table[$j - $S[$i]];
return $table[$n];
}
// Driver Code
$arr = array(1, 2, 3);
$m = sizeof($arr);
$n = 4;
$x = count_1($arr, $m, $n);
echo $x;
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// Dynamic Programming Javascript implementation
// of Coin Change problem
function count(S, m, n)
{
// table[i] will be storing the
// number of solutions for value i.
// We need n+1 rows as the table
// is constructed in bottom up manner
// using the base case (n = 0)
let table = new Array(n + 1);
table.fill(0);
// Base case (If given value is 0)
table[0] = 1;
// Pick all coins one by one and
// update the table[] values after
// the index greater than or equal
// to the value of the picked coin
for(let i = 0; i < m; i++)
for(let j = S[i]; j <= n; j++)
table[j] += table[j - S[i]];
return table[n];
}
let arr = [1, 2, 3];
let m = arr.length;
let n = 4;
document.write(count(arr, m, n));
</script>
Producción:
4
Gracias a Rohan Laishram por sugerir esta versión con espacio optimizado.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Referencias:
http://www.algorithmist.com/index.php/Coin_Change
El siguiente es otro enfoque de DP de arriba hacia abajo que utiliza la memorización:
C++
#include <bits/stdc++.h>
using namespace std;
int coinchange(vector<int>& a, int v, int n,
vector<vector<int> >& dp)
{
if (v == 0)
return dp[n][v] = 1;
if (n == 0)
return 0;
if (dp[n][v] != -1)
return dp[n][v];
if (a[n - 1] <= v) {
// Either Pick this coin or not
return dp[n][v] = coinchange(a, v - a[n - 1], n, dp)
+ coinchange(a, v, n - 1, dp);
}
else // We have no option but to leave this coin
return dp[n][v] = coinchange(a, v, n - 1, dp);
}
int32_t main()
{
int tc = 1;
// cin >> tc;
while (tc--) {
int n, v;
n = 3, v = 4;
vector<int> a = { 1, 2, 3 };
vector<vector<int> > dp(n + 1,
vector<int>(v + 1, -1));
int res = coinchange(a, v, n, dp);
cout << res << endl;
}
}
// This Code is Contributed by
// Harshit Agrawal NITT
Java
// Java program for the above approach
import java.util.*;
class GFG {
static int coinchange(int[] a, int v, int n, int[][] dp)
{
if (v == 0)
return dp[n][v] = 1;
if (n == 0)
return 0;
if (dp[n][v] != -1)
return dp[n][v];
if (a[n - 1] <= v)
{
// Either Pick this coin or not
return dp[n][v]
= coinchange(a, v - a[n - 1], n, dp)
+ coinchange(a, v, n - 1, dp);
}
else // We have no option but to leave this coin
return dp[n][v] = coinchange(a, v, n - 1, dp);
}
// Driver code
public static void main(String[] args)
{
int tc = 1;
while (tc != 0) {
int n, v;
n = 3;
v = 4;
int[] a = { 1, 2, 3 };
int[][] dp = new int[n + 1][v + 1];
for (int[] row : dp)
Arrays.fill(row, -1);
int res = coinchange(a, v, n, dp);
System.out.println(res);
tc--;
}
}
}
// This code is contributed by rajsanghavi9.
Python3
# Python program for the above approach def coinchange(a, v, n, dp): if (v == 0): dp[n][v] = 1; return dp[n][v]; if (n == 0): return 0; if (dp[n][v] != -1): return dp[n][v]; if (a[n - 1] <= v): # Either Pick this coin or not dp[n][v] = coinchange(a, v - a[n - 1], n, dp) + coinchange(a, v, n - 1, dp); return dp[n][v]; else: # We have no option but to leave this coin dp[n][v] = coinchange(a, v, n - 1, dp); return dp[n][v]; # Driver code if __name__ == '__main__': tc = 1; while (tc != 0): n = 3; v = 4; a = [ 1, 2, 3 ]; dp = [[-1 for i in range(v+1)] for j in range(n+1)] res = coinchange(a, v, n, dp); print(res); tc -= 1; # This code is contributed by Rajput-Ji
C#
// C# program for the above approach
using System;
public class GFG {
static int coinchange(int[] a, int v, int n, int[, ] dp)
{
if (v == 0)
return dp[n, v] = 1;
if (n == 0)
return 0;
if (dp[n, v] != -1)
return dp[n, v];
if (a[n - 1] <= v) {
// Either Pick this coin or not
return dp[n, v]
= coinchange(a, v - a[n - 1], n, dp)
+ coinchange(a, v, n - 1, dp);
}
else // We have no option but to leave this coin
return dp[n, v] = coinchange(a, v, n - 1, dp);
}
// Driver code
public static void Main(String[] args)
{
int tc = 1;
while (tc != 0) {
int n, v;
n = 3;
v = 4;
int[] a = { 1, 2, 3 };
int[, ] dp = new int[n + 1, v + 1];
for (int j = 0; j < n + 1; j++) {
for (int l = 0; l < v + 1; l++)
dp[j, l] = -1;
}
int res = coinchange(a, v, n, dp);
Console.WriteLine(res);
tc--;
}
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// javascript program for the above approach
function coinchange(a , v , n, dp) {
if (v == 0)
return dp[n][v] = 1;
if (n == 0)
return 0;
if (dp[n][v] != -1)
return dp[n][v];
if (a[n - 1] <= v) {
// Either Pick this coin or not
return dp[n][v] = coinchange(a, v - a[n - 1], n, dp) + coinchange(a, v, n - 1, dp);
} else // We have no option but to leave this coin
return dp[n][v] = coinchange(a, v, n - 1, dp);
}
// Driver code
var tc = 1;
while (tc != 0) {
var n, v;
n = 3;
v = 4;
var a = [ 1, 2, 3 ];
var dp = Array(n+1).fill().map(() => Array(v+1).fill(-1));
var res = coinchange(a, v, n, dp);
document.write(res);
tc--;
}
// This code contributed by umadevi9616
</script>
Producción
4
Complejidad temporal: O(M*N)
Espacio auxiliar: O(M*N)
Contribuido por: Mayukh Sinha
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA