Given an array arr[], and an integer N. The task is to maximize the sum of minimum and maximum of each group in a distribution of the array elements in N groups where the size of each group is given in an array b[] of size N.
Examples:
Input: a[] = {17, 12, 11, 9, 8, 8, 5, 4, 3}, N = 3,
b[] = {2, 3, 4}
Output: 60
Explanation: The array elements should be distributed in groups {17, 9} {12, 8, 8} {11, 5, 4, 3}.
So the sum becomes (17 + 9) + (12 + 8) + (11 + 3) = 26 + 20 + 14 = 60Input: a[] = {12, 3, 4, 2, 5, 9, 8, 1, 2}, N = 3,
b[] = {1, 4, 4}
Output: 45
Approach: This problem can be solved by using the Greedy Approach and some implementation. Follow the steps below to solve the given problem.
- sort array arr[] in descending order and b[] in ascending order.
- Initialize a variable ans to store the output.
- Iterate from i = 0 to i = N-1 in array a[] and add each element to ans.
- Initialize a variable ind to store the index of elements of the array arr[]. Assign N to variable ind.
- Take a loop from i=0 to N-1.
- If b[i] > 0 then increment ind with (b[i]-1)
- Add arr[ind] to ans, as arr[ind] will be the smallest element of that group
- increment ind with 1.
- output the ans.
See the illustration below for better understanding.
Illustration:
Consider an example: arr[] = {17, 12, 11, 9, 8, 8, 5, 4, 3}, N = 3 and b[] = {2, 3, 4}
Firstly, sort array arr[] in descending order and b[] in ascending order, After then put the first N greatest element of array arr[] to each group as shown in fig.
Secondly, Fill each group with the rest of the element of array arr[] (one group at a time)
Therefore answer will contain the sum of the first N elements of array arr[] i.e. 17, 12, 11 and also the last element which is filled in each group i.e. 9, 8 and 3.
Below is the implementation of the above approach.
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to maximum possible sum of minimum// and maximum elements of each groupint neveropen(int a[], int b[], int n, int k){ // Sorting array a in descending order // and array a in ascending order. sort(a, a + n, greater<int>()); sort(b, b + k); // Variable to store the required output. int ans = 0; // Loop to store sum of first k greatest // element of array a[] in ans. // since they will be gretest element // of each group when distributed // in group one by one. for (int i = 0; i < k; i++) { if (b[i] == 1) { ans += 2 * a[i]; } else { ans += a[i]; } --b[i]; } // Variable to store index of element a . int ind = k; // Then after when each grouped is filled, // then add a[ind] to ans as a[ind] will be // lowest element of each group. for (int i = 0; i < k; i++) { if (b[i] > 0) { ind += b[i] - 1; ans += a[ind]; ind++; } } return ans;}// Driver codeint main(){ int N = 3; // Size of array a[] int siz = 9; int a[] = { 17, 12, 11, 9, 8, 8, 5, 4, 3 }; int b[] = { 2, 3, 4 }; // Function Call cout << neveropen(a, b, 9, N);} |
Java
// Java code to find the maximum median// of a sub array having length at least K.import java.util.*;public class GFG{// Utility function to sort an array in// descending orderstatic void sort(int arr[]){ for (int i = 0; i < arr.length; i++) { for (int j = i + 1; j < arr.length; j++) { if(arr[i] < arr[j]) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } }// Function to maximum possible sum of minimum// and maximum elements of each groupstatic int neveropen(int a[], int b[], int n, int k){ // Sorting array a in descending order // and array b in ascending order. sort(a); Arrays.sort(b); // Variable to store the required output. int ans = 0; // Loop to store sum of first k greatest // element of array a[] in ans. // since they will be gretest element // of each group when distributed // in group one by one. for (int i = 0; i < k; i++) { if (b[i] == 1) { ans += 2 * a[i]; } else { ans += a[i]; } --b[i]; } // Variable to store index of element a . int ind = k; // Then after when each grouped is filled, // then add a[ind] to ans as a[ind] will be // lowest element of each group. for (int i = 0; i < k; i++) { if (b[i] > 0) { ind += b[i] - 1; ans += a[ind]; ind++; } } return ans;}// Driver codepublic static void main(String args[]){ int N = 3; // Size of array a[] int siz = 9; int a[] = { 17, 12, 11, 9, 8, 8, 5, 4, 3 }; int b[] = { 2, 3, 4 }; // Function Call System.out.println(neveropen(a, b, 9, N));}}// This code is contributed by Samim Hossain Mondal. |
Python
# Python program for the above approach# Function to maximum possible sum of minimum# and maximum elements of each groupdef neveropen(a, b, n, k): # Sorting array a in descending order # and array a in ascending order. a.sort(reverse=True) b.sort() # Variable to store the required output. ans = 0 # Loop to store sum of first k greatest # element of array a[] in ans. # since they will be gretest element # of each group when distributed # in group one by one. for i in range(0, k): if (b[i] == 1): ans = ans + (2 * a[i]) else: ans = ans + a[i] b[i] = b[i] - 1 # Variable to store index of element a . ind = k # Then after when each grouped is filled, # then add a[ind] to ans as a[ind] will be # lowest element of each group. for i in range(0, k): if (b[i] > 0): ind = ind + b[i] - 1 ans = ans + a[ind] ind = ind + 1 return ans# Driver codeN = 3# Size of array a[]siz = 9;a = [ 17, 12, 11, 9, 8, 8, 5, 4, 3 ]b = [ 2, 3, 4 ]# Function Callprint(neveropen(a, b, 9, N))# This code is contributed by Samim Hossain Mondal. |
C#
// C# code to find the maximum median// of a sub array having length at least K.using System;public class GFG{ // Utility function to sort an array in // descending order static void sort(int[] arr) { for (int i = 0; i < arr.Length; i++) { for (int j = i + 1; j < arr.Length; j++) { if(arr[i] < arr[j]) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } } // Function to maximum possible sum of minimum // and maximum elements of each group static int neveropen(int[] a, int[] b, int n, int k) { // Sorting array a in descending order // and array b in ascending order. sort(a); Array.Sort(b); // Variable to store the required output. int ans = 0; // Loop to store sum of first k greatest // element of array a[] in ans. // since they will be gretest element // of each group when distributed // in group one by one. for (int i = 0; i < k; i++) { if (b[i] == 1) { ans += 2 * a[i]; } else { ans += a[i]; } --b[i]; } // Variable to store index of element a . int ind = k; // Then after when each grouped is filled, // then add a[ind] to ans as a[ind] will be // lowest element of each group. for (int i = 0; i < k; i++) { if (b[i] > 0) { ind += b[i] - 1; ans += a[ind]; ind++; } } return ans; } // Driver code public static void Main() { int N = 3; // Size of array a[] int siz = 9; int[] a = { 17, 12, 11, 9, 8, 8, 5, 4, 3 }; int[] b = { 2, 3, 4 }; // Function Call Console.Write(neveropen(a, b, 9, N)); }}// This code is contributed by Saurabh Jaiswal |
Javascript
<script> // JavaScript code for the above approach // Function to maximum possible sum of minimum // and maximum elements of each group function neveropen(a, b, n, k) { // Sorting array a in descending order // and array a in ascending order. a.sort(function (a, b) { return b - a }) b.sort(function (a, b) { return a - b }) // Variable to store the required output. let ans = 0; // Loop to store sum of first k greatest // element of array a[] in ans. // since they will be gretest element // of each group when distributed // in group one by one. for (let i = 0; i < k; i++) { if (b[i] == 1) { ans += 2 * a[i]; } else { ans += a[i]; } --b[i]; } // Variable to store index of element a . let ind = k; // Then after when each grouped is filled, // then add a[ind] to ans as a[ind] will be // lowest element of each group. for (let i = 0; i < k; i++) { if (b[i] > 0) { ind += b[i] - 1; ans += a[ind]; ind++; } } return ans; } // Driver code let N = 3; // Size of array a[] let siz = 9; let a = [17, 12, 11, 9, 8, 8, 5, 4, 3]; let b = [2, 3, 4]; // Function Call document.write(neveropen(a, b, 9, N));// This code is contributed by Potta Lokesh </script> |
Output
60
Time complexity: O(M * logM) where M is the size of array arr.
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
