Minimize sum of smallest elements from K subsequences of length L

Given an array arr[] of size N, the task is to find the minimum possible sum by extracting the smallest element from any K subsequences from arr[] of length L such that each of the subsequences have no shared element. If it is not possible to get the required sum, print -1.
Examples: 
 

Input: arr[] = {2, 15, 5, 1, 35, 16, 67, 10}, K = 3, L = 2 
Output:
Explanation: 
Three subsequences of length 2 can be {1, 35}, {2, 15}, {5, 16} 
Minimum element of {1, 35} is 1. 
Minimum element of {2, 15} is 2. 
Minimum element of {5, 16} is 5. 
Their Sum is equal to 8 which is the minimum possible.
Input: arr[] = {19, 11, 21, 16, 22, 18, 14, 12}, K = 3, L = 3 
Output: -1 
Explanation: 
It is not possible to construct 3 subsequences of length 3 from arr[]. 
 

 

Approach: 
To optimize the above approach, we need to observe the following details: 
 

  • The K smallest elements of the array contribute to finding the minimum sum of the smallest elements of K subsequences.
  • The length of the array must be greater than or equal to (K * L) in order to form K subsequences of length L.

Follow the steps below to solve the problem: 
 



  • Check if the size of the array arr[] is greater than equal to (K * L).
  • If so, sort the array arr[] and print the sum of the first K elements of the array after sorting.
  • Otherwise, return -1.

Below is the implementation of the above approach:

C++

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// C++ Program to find the minimum 
// possible sum of the smallest 
// elements from K subsequences 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the minimum sum 
int findMinSum(int arr[], int K, 
               int L, int size) 
  
    if (K * L > size) 
        return -1; 
  
    int minsum = 0; 
  
    // Sort the array 
    sort(arr, arr + size); 
  
    // Calculate sum of smallest 
    // K elements 
    for (int i = 0; i < K; i++) 
        minsum += arr[i]; 
  
    // Return the sum 
    return minsum; 
  
// Driver Code 
int main() 
    int arr[] = { 2, 15, 5, 1, 
                  35, 16, 67, 10 }; 
    int K = 3; 
    int L = 2; 
  
    int length = sizeof(arr) 
                / sizeof(arr[0]); 
  
    cout << findMinSum(arr, K, 
                       L, length); 
  
    return 0; 

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Java

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// Java program to find the minimum
// possible sum of the smallest
// elements from K subsequences
import java.util.Arrays;
  
class GFG{
  
// Function to find the minimum sum
static int findMinSum(int []arr, int K,
                      int L, int size)
{
    if (K * L > size)
        return -1;
  
    int minsum = 0;
  
    // Sort the array
    Arrays.sort(arr); 
  
    // Calculate sum of smallest
    // K elements
    for(int i = 0; i < K; i++)
        minsum += arr[i];
  
    // Return the sum
    return minsum;
}
  
// Driver Code
public static void main(String args[]) 
{
    int arr[] = { 2, 15, 5, 1,
                  35, 16, 67, 10 };
    int K = 3;
    int L = 2;
    int length = arr.length;
  
    System.out.print(findMinSum(arr, K,
                                L, length));
}
}
  
// This code is contributed by Ritik Bansal

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Python3

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# Python3 program to find the minimum
# possible sum of the smallest
# elements from K subsequences
  
# Function to find the minimum sum
  
  
def findMinSum(arr, K, L, size):
  
    if (K * L > size):
        return -1
  
    minsum = 0
  
    # Sort the array
    arr.sort()
  
    # Calculate sum of smallest
    # K elements
    for i in range(K):
        minsum += arr[i]
  
    # Return the sum
    return minsum
  
  
# Driver code
if __name__ == '__main__':
  
    arr = [2, 15, 5, 1,
           35, 16, 67, 10]
    K = 3
    L = 2
  
    length = len(arr)
  
    print(findMinSum(arr, K, L, length))
  
# This code is contributed by Shivam Singh

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Output

8

Time Complexity: O(N * log(N)) 
Space Complexity: O(1)
 

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