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Split array into equal length subsets with maximum sum of Kth largest element of each subset
  • Last Updated : 03 Dec, 2020

Given an array arr[] of size N, two positive integers M and K, the task is to partition the array into M equal length subsets such that the sum of the Kth largest element of all these subsets is maximum. If it is not possible to partition the array into M equal length subsets, then print -1.

Examples:

Input: arr[] = { 1, 2, 3, 4, 5, 6 }, M = 2, K = 2 
Output:
Explanation: 
Length of each subset = (N / M) = 3. 
Possible subsets from the array are: { {1, 3, 4}, {2, 5, 6} } 
Therefore, the sum of Kth largest element from each subset = 3 + 5 = 8

Input: arr[] = {11, 20, 2, 17}, M = 4, K = 1 
Output: 50

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:



  • If N is not divisible M, then print -1.
  • Initialize a variable, say maxSum, to store the maximum possible sum of Kth largest element of all the subsets of the array by partitioning the array into M equal length subset.
  • Sort the array in descending order.
  • Iterate over the range [1, M] using variable i and update maxSum += arr[i * K – 1].
  • Finally, print the value of maxSum.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum of Kth
// largest element of M equal length partition
int maximumKthLargestsumPart(int arr[], int N,
                             int M, int K)
{
    // Stores sum of K_th largest element
    // of equal length partitions
    int maxSum = 0;
 
    // If N is not
    // divisible by M
    if (N % M)
        return -1;
 
    // Stores length of
    // each partition
    int sz = (N / M);
 
    // If K is greater than
    // length of partition
    if (K > sz)
        return -1;
 
    // Sort array in
    // descending porder
    sort(arr, arr + N, greater<int>());
 
    // Traverse the array
    for (int i = 1; i <= M;
         i++) {
 
        // Update maxSum
        maxSum
            += arr[i * K - 1];
    }
 
    return maxSum;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int M = 2;
    int K = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maximumKthLargestsumPart(arr, N, M, K);
 
    return 0;
}

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Java

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// Java program to implement
// the above approach
import java.util.*;
import java.util.Collections;
 
class GFG{
      
// Function to find the maximum sum of Kth
// largest element of M equal length partition
static int maximumKthLargestsumPart(int[] arr, int N,
                                    int M, int K)
{
     
    // Stores sum of K_th largest element
    // of equal length partitions
    int maxSum = 0;
     
    // If N is not
    // divisible by M
    if (N % M != 0)
        return -1;
    
    // Stores length of
    // each partition
    int sz = (N / M);
    
    // If K is greater than
    // length of partition
    if (K > sz)
        return -1;
    
    // Sort array in
    // descending porder
    Arrays.sort(arr);
    int i, k, t;
     
    for(i = 0; i < N / 2; i++)
    {
        t = arr[i];
        arr[i] = arr[N - i - 1];
        arr[N - i - 1] = t;
    }
 
    // Traverse the array
    for(i = 1; i <= M; i++)
    {
         
        // Update maxSum
        maxSum += arr[i * K - 1];
    }
    return maxSum;
}
  
// Driver code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int M = 2;
    int K = 1;
    int N = arr.length;
    
    System.out.println(maximumKthLargestsumPart(
        arr, N, M, K));
}
}
  
// This code is contributed by sanjoy_62

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Python3

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# Python3 program to implement
# the above approach
 
# Function to find the maximum sum of Kth
# largest element of M equal length partition
def maximumKthLargestsumPart(arr, N, M, K):
     
    # Stores sum of K_th largest element
    # of equal length partitions
    maxSum = 0
 
    # If N is not
    # divisible by M
    if (N % M != 0):
        return -1
 
    # Stores length of
    # each partition
    sz = (N / M)
 
    # If K is greater than
    # length of partition
    if (K > sz):
        return -1
 
    # Sort array in
    # descending porder
    arr = sorted(arr)
 
    for i in range(0, N // 2):
        t = arr[i]
        arr[i] = arr[N - i - 1]
        arr[N - i - 1] = t
 
    # Traverse the array
    for i in range(1, M + 1):
         
        # Update maxSum
        maxSum += arr[i * K - 1]
 
    return maxSum
 
# Driver code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3, 4, 5, 6 ]
    M = 2
    K = 1
    N = len(arr)
     
    print(maximumKthLargestsumPart(arr, N, M, K))
 
# This code is contributed by Amit Katiyar

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C#

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// C# program to implement
// the above approach
using System;
 
class GFG{
   
// Function to find the maximum sum of Kth
// largest element of M equal length partition
static int maximumKthLargestsumPart(int[] arr, int N,
                                    int M, int K)
{
     
    // Stores sum of K_th largest element
    // of equal length partitions
    int maxSum = 0;
   
    // If N is not
    // divisible by M
    if (N % M != 0)
        return -1;
   
    // Stores length of
    // each partition
    int sz = (N / M);
   
    // If K is greater than
    // length of partition
    if (K > sz)
        return -1;
   
    // Sort array in
    // descending porder
    Array.Sort(arr);
    Array.Reverse(arr);
   
    // Traverse the array
    for(int i = 1; i <= M; i++)
    {
         
        // Update maxSum
        maxSum += arr[i * K - 1];
    }
    return maxSum;
}
 
// Driver code   
static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int M = 2;
    int K = 1;
    int N = arr.Length;
   
    Console.WriteLine(maximumKthLargestsumPart(
        arr, N, M, K));
}
}
 
// This code is contributed by divyeshrabadiya07

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Output: 

11

 

Time complexity: O(N * log(N))
Auxiliary space: O(1)

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