Minimum sum of medians of all possible K length subsequences of a sorted array

Given a sorted array arr[] consisting of N integers and a positive integer K(such that N%K is 0), the task is to find the minimum sum of the medians of all possible subsequences of size K such that each element belongs to only one subsequence.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 2
Output: 6
Explanation:
Consider the subsequences of size K as {1, 4}, {2, 5}, and {3, 6}.
The sum of medians of all the subsequences is (1 + 2 + 3) = 6 which is the minimum possible sum.

Input: K = 3, arr[] = {3, 11, 12, 22, 33, 35, 38, 67, 69, 71, 94, 99}, K = 3
Output: 135

Naive Approach: The given problem can be solved by generating all possible K-sized sorted subsequences and print the median of all those subsequences as the result.

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the Greedy Approach for the construction of all the subsequences. The idea is to select K/2 elements from starting of the array and K/2 elements from the ending of the array such that the median always occurs in the first part. Follow the steps below to solve the problem:

• Initialize a variable, say res that stores the resultant sum of medians.
• Initialize a variable, say T as N/K to store the number of subsequences required and a variable D as (K + 1)/2 to store the distance between the medians.
• Initialize a variable, say i as (D - 1) to store the index of the first median to be added to the result.
• Iterate until the value of i < N and T > 0, and perform the following steps:
  • Add the value of arr[i] to the variable res.
  • Increment the value of i by D to get the index of the next median.
  • Decrement the value of T by 1.
• After completing the above steps, print the value of res as the result.

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 minimum sum of
// all the medians of the K sized sorted
// arrays formed from the given array
void sumOfMedians(int arr[], int N,
                  int K)
{
    // Stores the distance between
    // the medians
    int selectMedian = (K + 1) / 2;

    // Stores the number of subsequences
    // required
    int totalArrays = N / K;

    // Stores the resultant sum
    int minSum = 0;

    // Iterate from start and add
    // all the medians
    int i = selectMedian - 1;
    while (i < N and totalArrays != 0) {

        // Add the value of arr[i]
        // to the variable minsum
        minSum = minSum + arr[i];

        // Increment i by select the
        // median to get the next
        // median index
        i = i + selectMedian;

        // Decrement the value of
        // totalArrays by 1
        totalArrays--;
    }

    // Print the resultant minimum sum
    cout << minSum;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(int);
    int K = 2;
    sumOfMedians(arr, N, K);

    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

public class GFG {

    // Function to find the minimum sum of
    // all the medians of the K sized sorted
    // arrays formed from the given array
    static void sumOfMedians(int arr[], int N, int K)
    {
        // Stores the distance between
        // the medians
        int selectMedian = (K + 1) / 2;

        // Stores the number of subsequences
        // required
        int totalArrays = N / K;

        // Stores the resultant sum
        int minSum = 0;

        // Iterate from start and add
        // all the medians
        int i = selectMedian - 1;
        while (i < N && totalArrays != 0) {

            // Add the value of arr[i]
            // to the variable minsum
            minSum = minSum + arr[i];

            // Increment i by select the
            // median to get the next
            // median index
            i = i + selectMedian;

            // Decrement the value of
            // totalArrays by 1
            totalArrays--;
        }

        // Print the resultant minimum sum
        System.out.println(minSum);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        int N = arr.length;
        int K = 2;
        sumOfMedians(arr, N, K);
    }
}

// This code is contributed by Kingash. 

# Python3 program for the above approach

# Function to find the minimum sum of
# all the medians of the K sized sorted
# arrays formed from the given array
def sumOfMedians(arr, N, K):

    # Stores the distance between
    # the medians
    selectMedian = (K + 1) // 2

    # Stores the number of subsequences
    # required
    totalArrays = N // K

    # Stores the resultant sum
    minSum = 0

    # Iterate from start and add
    # all the medians
    i = selectMedian - 1
    
    while (i < N and totalArrays != 0):

        # Add the value of arr[i]
        # to the variable minsum
        minSum = minSum + arr[i]

        # Increment i by select the
        # median to get the next
        # median index
        i = i + selectMedian

        # Decrement the value of
        # totalArrays by 1
        totalArrays -= 1

     # Print the resultant minimum sum
    print(minSum)

 # Driver Code
if __name__ == '__main__':
    
    arr = [ 1, 2, 3, 4, 5, 6 ]
    N = len(arr)
    K = 2
    
    sumOfMedians(arr, N, K)

# This code is contributed by nirajgsuain5

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
    // Function to find the minimum sum of
    // all the medians of the K sized sorted
    // arrays formed from the given array
    static void sumOfMedians(int[] arr, int N, int K)
    {
      
        // Stores the distance between
        // the medians
        int selectMedian = (K + 1) / 2;
 
        // Stores the number of subsequences
        // required
        int totalArrays = N / K;
 
        // Stores the resultant sum
        int minSum = 0;
 
        // Iterate from start and add
        // all the medians
        int i = selectMedian - 1;
        while (i < N && totalArrays != 0) {
 
            // Add the value of arr[i]
            // to the variable minsum
            minSum = minSum + arr[i];
 
            // Increment i by select the
            // median to get the next
            // median index
            i = i + selectMedian;
 
            // Decrement the value of
            // totalArrays by 1
            totalArrays--;
        }
 
        // Print the resultant minimum sum
        Console.WriteLine(minSum);
    }
 
// Driver Code
public static void Main()
{
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int N = arr.Length;
        int K = 2;
        sumOfMedians(arr, N, K);
     
}
}

// This code is contributed by code_hunt.

<script>

// JavaScript program for the above approach

    // Function to find the minimum sum of
    // all the medians of the K sized sorted
    // arrays formed from the given array
    function sumOfMedians(arr, N, K)
    {
        // Stores the distance between
        // the medians
        let selectMedian = Math.floor((K + 1) / 2);
 
        // Stores the number of subsequences
        // required
        let totalArrays =  Math.floor(N / K);
 
        // Stores the resultant sum
        let minSum = 0;
 
        // Iterate from start and add
        // all the medians
        let i = selectMedian - 1;
        while (i < N && totalArrays != 0) {
 
            // Add the value of arr[i]
            // to the variable minsum
            minSum = minSum + arr[i];
 
            // Increment i by select the
            // median to get the next
            // median index
            i = i + selectMedian;
 
            // Decrement the value of
            // totalArrays by 1
            totalArrays--;
        }
 
        // Print the resultant minimum sum
       document.write(minSum);
    }


// Driver Code

        let arr = [ 1, 2, 3, 4, 5, 6 ];
        let N = arr.length;
        let K = 2;
        sumOfMedians(arr, N, K);    

</script>

6

Time Complexity: O(N)
Auxiliary Space: O(1)