Given an array arr[] of size N and a positive integer K, the task is to find the minimum possible cost to split the array into K subsets, where the cost of ith element ( 1-based indexing ) of each subset is equal to the product of that element and i.
Examples:
Input: arr[] = { 2, 3, 4, 1 }, K = 3
Output: 11
Explanation:
Split the array arr[] into K(= 3) subsets { { 4, 1 }, { 2 }, { 3 } }
Total cost of 1st subset = 4 * 1 + 1 * 2 = 6
Total cost of 2nd subset = 2 * 1 = 2
Total cost of 3rd subset = 3 * 1 = 3
Therefore, the total cost of K(= 3) subsets is 6 + 2 + 3 = 11.Input: arr[] = { 9, 20, 7, 8 }, K=2
Output: 59
Explanation:
Dividing the array arr[] into K(= 3) subsets { { 20, 8 }, { 9, 7 } }
Total cost of 1st subset = 20 * 1 + 8 * 2 = 36
Total cost of 2nd subset = 9 * 1 + 7 * 2 = 23
Therefore, the total cost of K(= 3) subsets is 36 + 23 = 59
Approach: The problem can be solved using Greedy technique. The idea is to divide the array elements such all elements in respective subsets is in decreasing order. Follow the steps below to solve the problem:
- Sort the given array in descending order.
- Initialize a variable, say totalCost, to store the minimum cost to split the array into K subsets.
- Initialize a variable, say X, to store the position of an element in a subset.
- Iterate over the range [1, N] using variable i. For every ith operation, increment the value of totalCost by ((arr[i]+ …+ arr[i + K]) * X) and update i = i + K, X += 1.
- Finally, print the value of totalCost.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum cost to // split array into K subsets int getMinCost( int * arr, int n, int k)
{ // Sort the array in descending order
sort(arr, arr + n, greater< int >());
// Stores minimum cost to split
// the array into K subsets
int min_cost = 0;
// Stores position of
// elements of a subset
int X = 0;
// Iterate over the range [1, N]
for ( int i = 0; i < n; i += k) {
// Calculate the cost to select
// X-th element of every subset
for ( int j = i; j < i + k && j < n; j++) {
// Update min_cost
min_cost += arr[j] * (X + 1);
}
// Update X
X++;
}
return min_cost;
} // Driver Code int main()
{ int arr[] = { 9, 20, 7, 8 };
int K = 2;
int N = sizeof (arr)
/ sizeof (arr[0]);
// Function call
cout << getMinCost(arr, N, K) << endl;
} |
// Java program to implement // the above approach import java.util.*;
class GFG
{ // reverses an array static void reverse( int a[], int n)
{ int i, k, t;
for (i = 0 ; i < n / 2 ; i++)
{
t = a[i];
a[i] = a[n - i - 1 ];
a[n - i - 1 ] = t;
}
} // Function to find the minimum cost to // split array into K subsets static int getMinCost( int [] arr, int n, int k)
{ // Sort the array in descending order
Arrays.sort(arr);
reverse(arr, n);
// Stores minimum cost to split
// the array into K subsets
int min_cost = 0 ;
// Stores position of
// elements of a subset
int X = 0 ;
// Iterate over the range [1, N]
for ( int i = 0 ; i < n; i += k)
{
// Calculate the cost to select
// X-th element of every subset
for ( int j = i; j < i + k && j < n; j++)
{
// Update min_cost
min_cost += arr[j] * (X + 1 );
}
// Update X
X++;
}
return min_cost;
} // Driver code public static void main(String[] args)
{ int arr[] = { 9 , 20 , 7 , 8 };
int K = 2 ;
int N = arr.length;
// Function call
System.out.println( getMinCost(arr, N, K));
} } // This code is contributed by susmitakundugoaldanga |
# Python program to implement # the above approach # Function to find the minimum cost to # split array into K subsets def getMinCost(arr, n, k):
# Sort the array in descending order
arr.sort(reverse = True )
# Stores minimum cost to split
# the array into K subsets
min_cost = 0 ;
# Stores position of
# elements of a subset
X = 0 ;
# Iterate over the range [1, N]
for i in range ( 0 , n, k):
# Calculate the cost to select
# X-th element of every subset
for j in range (i, n, 1 ):
# Update min_cost
if (j < i + k):
min_cost + = arr[j] * (X + 1 );
# Update X
X + = 1 ;
return min_cost;
# Driver code if __name__ = = '__main__' :
arr = [ 9 , 20 , 7 , 8 ];
K = 2 ;
N = len (arr);
# Function call
print (getMinCost(arr, N, K));
# This code is contributed by 29AjayKumar |
// C# program to implement // the above approach using System;
class GFG
{ // reverses an array static void reverse( int []a, int n)
{ int i, k, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
} // Function to find the minimum cost to // split array into K subsets static int getMinCost( int [] arr, int n, int k)
{ // Sort the array in descending order
Array.Sort(arr);
reverse(arr, n);
// Stores minimum cost to split
// the array into K subsets
int min_cost = 0;
// Stores position of
// elements of a subset
int X = 0;
// Iterate over the range [1, N]
for ( int i = 0; i < n; i += k)
{
// Calculate the cost to select
// X-th element of every subset
for ( int j = i; j < i + k && j < n; j++)
{
// Update min_cost
min_cost += arr[j] * (X + 1);
}
// Update X
X++;
}
return min_cost;
} // Driver code public static void Main(String[] args)
{ int []arr = { 9, 20, 7, 8 };
int K = 2;
int N = arr.Length;
// Function call
Console.WriteLine( getMinCost(arr, N, K));
} } // This code is contributed by shikhasingrajput |
<script> // JavaScript program to implement // the above approach // Reverses an array function reverse(a, n)
{ var i, k, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
} // Function to find the minimum cost to // split array into K subsets function getMinCost(arr, n, k)
{ // Sort the array in descending order
arr.sort((a, b) => b - a);
// Stores minimum cost to split
// the array into K subsets
var min_cost = 0;
// Stores position of
// elements of a subset
var X = 0;
// Iterate over the range [1, N]
for ( var i = 0; i < n; i += k)
{
// Calculate the cost to select
// X-th element of every subset
for ( var j = i; j < i + k && j < n; j++)
{
// Update min_cost
min_cost += arr[j] * (X + 1);
}
// Update X
X++;
}
return min_cost;
} // Driver code var arr = [ 9, 20, 7, 8 ];
var K = 2;
var N = arr.length;
// Function call document.write(getMinCost(arr, N, K)); // This code is contributed by rdtank </script> |
59
Time Complexity: O(N * log(N))
Auxiliary Space: O(1) it is using constant space for variables