Dada una array arr[] y un entero K . La tarea es dividir el arreglo en K partes (subarreglo) de tal manera que la suma de los valores de todos los subarreglos sea mínima.
El valor de cada subarreglo se define como:
- Tome el máximo de ese subarreglo.
- Resta cada elemento del subarreglo con el máximo.
- Tome la suma de todos los valores después de la resta.
La tarea es minimizar la suma de los valores después de dividir la array en K partes.
Ejemplos:
Entrada: arr[] = { 2, 9, 5, 4, 8, 3, 6 }, K = 2
Salida: 19
Explicación:
Los dos grupos son: {2} con max = 2 y {9, 5, 4, 8, 3, 6} con max=9,
suma de la diferencia del primer grupo = 2 – 2 = 0,
suma de la diferencia del segundo grupo = (9-9) + (9-5) + (9-4) + ( 9-8) + (9-3) + (9-6) = 19
Entrada: arr[] = { 12, 20, 30, 14, 25}, K = 3
Salida: 19
Enfoque:
La solución de fuerza bruta será probar todas las particiones posibles y tomar el mínimo total. Aunque esta solución es exponencial en el tiempo. En la solución recursiva, hay muchos subproblemas superpuestos que pueden optimizarse mediante programación dinámica.
Entonces, podemos formar una fórmula recursiva básica y que calcule todas las soluciones posibles y encuentre la mejor solución posible. Podemos ver que la solución recursiva tiene muchos subproblemas superpuestos, podemos reducir la complejidad usando la programación dinámica.
Fórmula recursiva:
F(i, K) = { min de todos los valores tales que j < i [ max(Arr[i..j]) * (i – j + 1) – Sum(A[i…j] ] } + F(j, K-1)
El enfoque ascendente se puede utilizar para calcular primero los valores de los subproblemas y almacenarlos.Aquí
dp [i][j]define el valor mínimo que se puede obtener si la array parte del índice i y tiene una partición j .
Entonces, la respuesta a los problemas será dp[0][K] , una array que comienza en 0 y tiene K particiones.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to divide an array into k
// parts such that the sum of difference
// of every element with the maximum element
// of that part is minimum
int divideArray(int arr[], int n, int k)
{
// Dp to store the values
int dp[500][500] = { 0 };
k -= 1;
// Fill up the dp table
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j <= k; j++) {
// Intitilize maximum value
dp[i][j] = INT_MAX;
// Max element and the sum
int max_ = -1, sum = 0;
// Run a loop from i to n
for (int l = i; l < n; l++) {
// Find the maximum number
// from i to l and the sum
// from i to l
max_ = max(max_, arr[l]);
sum += arr[l];
// Find the sum of difference
// of every element with the
// maximum element
int diff = (l - i + 1) * max_ - sum;
// If the array can be divided
if (j > 0)
dp[i][j]
= min(dp[i][j],
diff + dp[l + 1][j - 1]);
else
dp[i][j] = diff;
}
}
}
// Returns the minimum sum
// in K parts
return dp[0][k];
}
// Driver code
int main()
{
int arr[] = { 2, 9, 5, 4, 8, 3, 6 };
int n = sizeof(arr) / sizeof(int);
int k = 2;
cout << divideArray(arr, n, k) << "\n";
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to divide an array into k
// parts such that the sum of difference
// of every element with the maximum element
// of that part is minimum
static int divideArray(int arr[], int n, int k)
{
// Dp to store the values
int dp[][] = new int[500][500];
int i, j;
for(i = 0; i < 500; i++)
for(j = 0; j < 500; j++)
dp[i][j] = 0;
k -= 1;
// Fill up the dp table
for (i = n - 1; i >= 0; i--)
{
for (j = 0; j <= k; j++)
{
// Intitilize maximum value
dp[i][j] = Integer.MAX_VALUE;
// Max element and the sum
int max_ = -1, sum = 0;
// Run a loop from i to n
for (int l = i; l < n; l++)
{
// Find the maximum number
// from i to l and the sum
// from i to l
max_ = Math.max(max_, arr[l]);
sum += arr[l];
// Find the sum of difference
// of every element with the
// maximum element
int diff = (l - i + 1) * max_ - sum;
// If the array can be divided
if (j > 0)
dp[i][j] = Math.min(dp[i][j], diff +
dp[l + 1][j - 1]);
else
dp[i][j] = diff;
}
}
}
// Returns the minimum sum
// in K parts
return dp[0][k];
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 2, 9, 5, 4, 8, 3, 6 };
int n = arr.length;
int k = 2;
System.out.println(divideArray(arr, n, k));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the above approach # Function to divide an array into k # parts such that the summ of difference # of every element with the maximum element # of that part is minimum def divideArray(arr, n, k): # Dp to store the values dp = [[0 for i in range(500)] for i in range(500)] k -= 1 # Fill up the dp table for i in range(n - 1, -1, -1): for j in range(0, k + 1): # Intitilize maximum value dp[i][j] = 10**9 # Max element and the summ max_ = -1 summ = 0 # Run a loop from i to n for l in range(i, n): # Find the maximum number # from i to l and the summ # from i to l max_ = max(max_, arr[l]) summ += arr[l] # Find the summ of difference # of every element with the # maximum element diff = (l - i + 1) * max_ - summ # If the array can be divided if (j > 0): dp[i][j]= min(dp[i][j], diff + dp[l + 1][j - 1]) else: dp[i][j] = diff # Returns the minimum summ # in K parts return dp[0][k] # Driver code arr = [2, 9, 5, 4, 8, 3, 6] n = len(arr) k = 2 print(divideArray(arr, n, k)) # This code is contributed by Mohit Kumar
C#
// C# implementation of above approach
using System;
class GFG
{
// Function to divide an array into k
// parts such that the sum of difference
// of every element with the maximum element
// of that part is minimum
static int divideArray(int []arr, int n, int k)
{
// Dp to store the values
int [,]dp = new int[500, 500];
int i, j;
for(i = 0; i < 500; i++)
for(j = 0; j < 500; j++)
dp[i, j] = 0;
k -= 1;
// Fill up the dp table
for (i = n - 1; i >= 0; i--)
{
for (j = 0; j <= k; j++)
{
// Intitilize maximum value
dp[i, j] = int.MaxValue;
// Max element and the sum
int max_ = -1, sum = 0;
// Run a loop from i to n
for (int l = i; l < n; l++)
{
// Find the maximum number
// from i to l and the sum
// from i to l
max_ = Math.Max(max_, arr[l]);
sum += arr[l];
// Find the sum of difference
// of every element with the
// maximum element
int diff = (l - i + 1) * max_ - sum;
// If the array can be divided
if (j > 0)
dp[i, j] = Math.Min(dp[i, j], diff +
dp[l + 1, j - 1]);
else
dp[i, j] = diff;
}
}
}
// Returns the minimum sum
// in K parts
return dp[0, k];
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 2, 9, 5, 4, 8, 3, 6 };
int n = arr.Length;
int k = 2;
Console.WriteLine(divideArray(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation of the above approach
// Function to divide an array into k
// parts such that the sum of difference
// of every element with the maximum element
// of that part is minimum
function divideArray(arr, n, k)
{
// Dp to store the values
var dp = Array.from(Array(500), ()=> Array(500).fill(0));
k -= 1;
// Fill up the dp table
for (var i = n - 1; i >= 0; i--) {
for (var j = 0; j <= k; j++) {
// Intitilize maximum value
dp[i][j] = 1000000000;
// Max element and the sum
var max_ = -1, sum = 0;
// Run a loop from i to n
for (var l = i; l < n; l++) {
// Find the maximum number
// from i to l and the sum
// from i to l
max_ = Math.max(max_, arr[l]);
sum += arr[l];
// Find the sum of difference
// of every element with the
// maximum element
var diff = (l - i + 1) * max_ - sum;
// If the array can be divided
if (j > 0)
dp[i][j]
= Math.min(dp[i][j],
diff + dp[l + 1][j - 1]);
else
dp[i][j] = diff;
}
}
}
// Returns the minimum sum
// in K parts
return dp[0][k];
}
// Driver code
var arr = [2, 9, 5, 4, 8, 3, 6 ];
var n = arr.length;
var k = 2;
document.write( divideArray(arr, n, k) + "<br>");
</script>
19
Complejidad de tiempo: O(n*n*k), n es el tamaño de la array
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por andrew1234 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA