Given an array A[] of size N and a positive integer K ( which will always be a factor of N), the task is to find the maximum possible sum of the second smallest elements of each partition of the array by partitioning the array into (N / K) partitions of equal size.
Examples:
Input: A[] = {2, 3, 1, 4, 7, 5, 6, 1}, K = 4
Output: 7
Explanation: Split the array as {1, 2, 3, 4} and {1, 5, 6, 7}. Therefore, sum = 2 + 5 = 7, which is the maximum possible sum.Input: A[] = {12, 43, 15, 32, 45, 23}, K = 3
Output : 66
Explanation: Split the array as {12, 23, 32} and {15, 43, 45}. Therefore, sum = 23 + 43 = 66, which is the maximum possible sum.
Approach: The idea is to sort the given array in ascending order and in order to maximize the required sum, divide the first N / K elements of A[] to each of the arrays as their first term, then choose every (K – 1)th element of A[] starting from N/K.
Follow the steps below to solve the problem:
- Sort the array A[] in increasing order.
- Initialize sum with 0 to store the required sum.
- Now, initialize a variable i with N / K.
- While i is less than N, perform the following steps:
- Increment sum by A[i].
- Increment i by K – 1.
- After traversing, print sum as the required answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum sum of // second smallest of each partition // of size K void findSum( int A[], int N, int K)
{ // Sort the array A[]
// in ascending order
sort(A, A + N);
// Store the maximum sum of second
// smallest of each partition
// of size K
int sum = 0;
// Select every (K-1)th element as
// second smallest element
for ( int i = N / K; i < N; i += K - 1) {
// Update sum
sum += A[i];
}
// Print the maximum sum
cout << sum;
} // Driver Code int main()
{ // Given size of partitions
int K = 4;
// Given array A[]
int A[] = { 2, 3, 1, 4, 7, 5, 6, 1 };
// Size of the given array
int N = sizeof (A) / sizeof (A[0]);
// Function Call
findSum(A, N, K);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the maximum sum of // second smallest of each partition // of size K static void findSum( int A[], int N, int K)
{ // Sort the array A[]
// in ascending order
Arrays.sort(A);
// Store the maximum sum of second
// smallest of each partition
// of size K
int sum = 0 ;
// Select every (K-1)th element as
// second smallest element
for ( int i = N / K; i < N; i += K - 1 )
{
// Update sum
sum += A[i];
}
// Print the maximum sum
System.out.print(sum);
} // Driver Code public static void main(String[] args)
{ // Given size of partitions
int K = 4 ;
// Given array A[]
int A[] = { 2 , 3 , 1 , 4 , 7 , 5 , 6 , 1 };
// Size of the given array
int N = A.length;
// Function Call
findSum(A, N, K);
} } // This code is contributed by shikhasingrajput |
# Python3 program for the above approach # Function to find the maximum sum of # second smallest of each partition # of size K def findSum(A, N, K):
# Sort the array A
# in ascending order
A.sort();
# Store the maximum sum of second
# smallest of each partition
# of size K
sum = 0 ;
# Select every (K-1)th element as
# second smallest element
for i in range (N / / K, N, K - 1 ):
# Update sum
sum + = A[i];
# Print the maximum sum
print ( sum );
# Driver Code if __name__ = = '__main__' :
# Given size of partitions
K = 4 ;
# Given array A
A = [ 2 , 3 , 1 , 4 , 7 , 5 , 6 , 1 ];
# Size of the given array
N = len (A);
# Function Call
findSum(A, N, K);
# This code contributed by shikhasingrajput
|
// C# program for the above approach using System;
class GFG{
// Function to find the maximum sum of // second smallest of each partition // of size K static void findSum( int []A, int N, int K)
{ // Sort the array []A
// in ascending order
Array.Sort(A);
// Store the maximum sum of second
// smallest of each partition
// of size K
int sum = 0;
// Select every (K-1)th element as
// second smallest element
for ( int i = N / K; i < N; i += K - 1)
{
// Update sum
sum += A[i];
}
// Print the maximum sum
Console.Write(sum);
} // Driver Code public static void Main(String[] args)
{ // Given size of partitions
int K = 4;
// Given array []A
int []A = { 2, 3, 1, 4, 7, 5, 6, 1 };
// Size of the given array
int N = A.Length;
// Function Call
findSum(A, N, K);
} } // This code is contributed by shikhasingrajput |
<script> // Javascript program of the above approach // Function to find the maximum sum of // second smallest of each partition // of size K function findSum(A, N, K)
{ // Sort the array A[]
// in ascending order
A.sort();
// Store the maximum sum of second
// smallest of each partition
// of size K
let sum = 0;
// Select every (K-1)th element as
// second smallest element
for (let i = N / K; i < N; i += K - 1)
{
// Update sum
sum += A[i];
}
// Print the maximum sum
document.write(sum);
} // Driver Code // Given size of partitions let K = 4; // Given array A[] let A = [ 2, 3, 1, 4, 7, 5, 6, 1 ]; // Size of the given array let N = A.length; // Function Call findSum(A, N, K); // This code is contributed by chinmoy1997pal </script> |
7
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)