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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that find the minimum cost of // selecting array elementvoid minimumCost(int arr[], int N, int M) { // Sorting the given array in // increasing order sort(arr, arr + N); // To store the prefix sum of arr[] int pref[N]; pref[0] = arr[0]; for(int i = 1; i < N; i++) { pref[i] = arr[i] + pref[i-1]; } // Update the pref[] to find the cost // selecting array element by selecting // at most M element for(int i = M; i < N; i++) { pref[i] += pref[i-M]; } // Print the pref[] for the result for(int i = 0; i < N; i++) { cout << pref[i] << ' '; } }// Driver Codeint 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
// Java program for the above approachimport java.util.*;class GFG{// Function that find the minimum cost of // selecting array elementstatic void minimumCost(int arr[], int N, int M) { // Sorting the given array in // increasing order Arrays.sort(arr); // To store the prefix sum of arr[] int []pref = new int[N]; pref[0] = arr[0]; for(int i = 1; i < N; i++) { pref[i] = arr[i] + pref[i - 1]; } // Update the pref[] to find the cost // selecting array element by selecting // at most M element for(int i = M; i < N; i++) { pref[i] += pref[i - M]; } // Print the pref[] for the result for(int i = 0; i < N; i++) { System.out.print(pref[i] + " "); }}// Driver Codepublic 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);}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 program for the above approach # Function that find the minimum cost # of selecting array element def minimumCost(arr, N, M): # Sorting the given array in # increasing order arr.sort() # To store the prefix sum of arr[] pref = [] pref.append(arr[0]) for i in range(1, N): pref.append(arr[i] + pref[i - 1]) # Update the pref[] to find the cost # selecting array element by selecting # at most M element for i in range(M, N): pref[i] += pref[i - M] # Print the pref[] for the result for i in range(N): print(pref[i], end = ' ')# Driver Code arr = [ 6, 19, 3, 4, 4, 2, 6, 7, 8 ]M = 2N = len(arr)minimumCost(arr, N, M);# This code is contributed by yatinagg |
C#
// C# program for the above approachusing System;class GFG{// Function that find the minimum cost // of selecting array elementstatic void minimumCost(int []arr, int N, int M) { // Sorting the given array // in increasing order Array.Sort(arr); // To store the prefix sum of []arr int []pref = new int[N]; pref[0] = arr[0]; for(int i = 1; i < N; i++) { pref[i] = arr[i] + pref[i - 1]; } // Update the pref[] to find the cost // selecting array element by selecting // at most M element for(int i = M; i < N; i++) { pref[i] += pref[i - M]; } // Print the pref[] for the result for(int i = 0; i < N; i++) { Console.Write(pref[i] + " "); }}// Driver Codepublic 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);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Function that find the minimum cost of // selecting array elementfunction minimumCost(arr, N, M) { // Sorting the given array in // increasing order arr.sort((a, b) => a - b); // To store the prefix sum of arr[] 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]; } // Update the pref[] to find the cost // selecting array element by selecting // at most M element for(let i = M; i < N; i++) { pref[i] += pref[i - M]; } // Print the pref[] for the result for(let i = 0; i < N; i++) { document.write(pref[i] + " "); }}// Driver Code 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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