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(2N)
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)