从长度为L的K个子序列中最小化最小元素的总和

给定大小为N的数组arr [] ，任务是通过从长度L的arr []的任何K个子序列中提取最小的元素来找到最小的总和，以使每个子序列都没有共享元素。如果无法获得所需的总和，请打印-1。

例子：

Input: arr[] = {2, 15, 5, 1, 35, 16, 67, 10}, K = 3, L = 2 
Output: 8 
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[]. 

方法：
为了优化上述方法，我们需要注意以下细节：

• 数组的K个最小元素有助于找到K个子序列的最小元素的最小和。
• 数组的长度必须大于或等于(K * L)，才能形成长度为L的K个子序列。

请按照以下步骤解决问题：

• 检查数组arr []的大小是否大于(K * L) 。
• 如果是这样，请对数组arr []进行排序，并在排序后打印该数组的前K个元素的总和。
• 否则，返回-1。

下面是上述方法的实现：

C++

// C++ Program to find the minimum
// possible sum of the smallest
// elements from K subsequences
 
#include 
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;
}

Java

// 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

Python3

# 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

C#

// C# program to find the minimum
// possible sum of the smallest
// elements from K subsequences
using System;
   
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
    Array.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() 
{
    int[] arr = { 2, 15, 5, 1,
                  35, 16, 67, 10 };
    int K = 3;
    int L = 2;
    int length = arr.Length;
   
    Console.Write(findMinSum(arr, K,
                             L, length));
}
}
 
// This code is contributed by code_hunt

输出

8

时间复杂度： O(N * log(N))
空间复杂度： O(1)