Minimum cost of choosing the array element
Last Updated :
12 Apr, 2023
Given an array arr[] of N integers and an integer M and the cost of selecting any array element(say x) at any day(say d), is x*d. The task is to minimize the cost of selecting 1, 2, 3, …, N array where each day at most M elements is allowed to select.
Examples:
Input: arr[] = {6, 19, 3, 4, 4, 2, 6, 7, 8}, M = 2
Output: 2 5 11 18 30 43 62 83 121
Explanation:
For selecting 1, 2, 3, .. , N elements when at most 2 elements are allowed to select each day:
The Cost of selecting 1 element:
select one smallest element on day 1, then cost is 2*1 = 2
The Cost of selecting 2 elements:
select two smallest elements on day 1, then cost is (2+3)*1 = 5
The Cost of selecting 3 elements:
select 2nd and 3rd smallest elements on day 1, then cost is (3+4)*1 = 7
select 1st smallest element on day 2, then cost is 2*2 = 4
So, the total cost is 7 + 4 = 11
Similarly, we can find the cost for selecting 4, 5, 6, 7, 8 and 9 elements is 18, 30, 43, 62, 83 and 121 respectively.
Input: arr[] = {6, 19, 12, 6, 7, 9}, M = 3
Output: 6 12 19 34 52 78
Approach: The idea is to use Prefix Sum Array.
- Sort the given array in increasing order.
- Store the prefix sum of the sorted array in pref[]. This prefix sum gives the minimum cost of selecting the 1, 2, 3, … N array elements when atmost one element is allowed to select each day.
- To find the minimum cost when atmost M element is allowed to select each day, update the prefix array pref[] from index M to N as:
pref[i] = pref[i] + pref[i-M]
- For Example:
arr[] = {6, 9, 3, 4, 4, 2, 6, 7, 8}
After sorting arr[]:
arr[] = {2, 3, 4, 4, 6, 6, 7, 8, 9}
Prefix array is:
pref[] = {2, 5, 9, 13, 19, 25, 32, 40, 49}
Now at every index i, pref[i] gives the cost
of selecting i array element when atmost one
element is allowed to select each day.
-
Now for M = 3, when at most 3 elements
are allowed to select each day, then
by update every index(from M to N)
of pref[] as:
pref[i] = pref[i] + pref[i-M]
the cost of selecting elements
from (i-M+1)th to ith index on day 1,
the cost of selecting elements
from (i-M)th to (i-2*M)th index on day 2
...
...
...
the cost of selecting elements
from (i-n*M)th to 0th index on day N.
-
- After the above step, every index(say i) of prefix array pref[] stores the cost selecting i elements when atmost M elements are allowed to select each day.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void minimumCost( int arr[], int N, int M) {
sort(arr, arr + N);
int pref[N];
pref[0] = arr[0];
for ( int i = 1; i < N; i++) {
pref[i] = arr[i] + pref[i-1];
}
for ( int i = M; i < N; i++) {
pref[i] += pref[i-M];
}
for ( int i = 0; i < N; i++) {
cout << pref[i] << ' ' ;
}
}
int main()
{
int arr[] = {6, 19, 3, 4, 4, 2, 6, 7, 8};
int M = 2;
int N = sizeof (arr)/ sizeof (arr[0]);
minimumCost(arr, N, M);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void minimumCost( int arr[], int N, int M)
{
Arrays.sort(arr);
int []pref = new int [N];
pref[ 0 ] = arr[ 0 ];
for ( int i = 1 ; i < N; i++)
{
pref[i] = arr[i] + pref[i - 1 ];
}
for ( int i = M; i < N; i++)
{
pref[i] += pref[i - M];
}
for ( int i = 0 ; i < N; i++)
{
System.out.print(pref[i] + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 6 , 19 , 3 , 4 , 4 , 2 , 6 , 7 , 8 };
int M = 2 ;
int N = arr.length;
minimumCost(arr, N, M);
}
}
|
Python3
def minimumCost(arr, N, M):
arr.sort()
pref = []
pref.append(arr[ 0 ])
for i in range ( 1 , N):
pref.append(arr[i] + pref[i - 1 ])
for i in range (M, N):
pref[i] + = pref[i - M]
for i in range (N):
print (pref[i], end = ' ' )
arr = [ 6 , 19 , 3 , 4 , 4 , 2 , 6 , 7 , 8 ]
M = 2
N = len (arr)
minimumCost(arr, N, M);
|
C#
using System;
class GFG{
static void minimumCost( int []arr, int N,
int M)
{
Array.Sort(arr);
int []pref = new int [N];
pref[0] = arr[0];
for ( int i = 1; i < N; i++)
{
pref[i] = arr[i] + pref[i - 1];
}
for ( int i = M; i < N; i++)
{
pref[i] += pref[i - M];
}
for ( int i = 0; i < N; i++)
{
Console.Write(pref[i] + " " );
}
}
public static void Main(String[] args)
{
int []arr = { 6, 19, 3, 4, 4,
2, 6, 7, 8 };
int M = 2;
int N = arr.Length;
minimumCost(arr, N, M);
}
}
|
Javascript
<script>
function minimumCost(arr, N, M)
{
arr.sort((a, b) => a - b);
let pref = Array.from({length: N}, (_, i) => 0);
pref[0] = arr[0];
for (let i = 1; i < N; i++)
{
pref[i] = arr[i] + pref[i - 1];
}
for (let i = M; i < N; i++)
{
pref[i] += pref[i - M];
}
for (let i = 0; i < N; i++)
{
document.write(pref[i] + " " );
}
}
let arr = [ 6, 19, 3, 4, 4, 2, 6, 7, 8 ];
let M = 2;
let N = arr.length;
minimumCost(arr, N, M);
</script>
|
Output:
2 5 11 18 30 43 62 83 121
Time Complexity: O(N*log N), where N is the number of element in the array.
Space Complexity: O(N) as pref array has been created. Here, N is the number of element in the array.
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