从串联M次的数组中找出第K个最小元素

给定一个数组arr []和两个整数K和M。问题是将数组与自身连接M次后，找到第K个极小值元素。

例子：

Input  : arr[] = {3, 1, 2}, K = 4, M = 3 
Output : 4'th Minimum element is : 2
Explanation: Concatenate array 3 times (ie., M = 3)
              arr[] = [3, 1, 2, 3, 1, 2, 3, 1, 2]
              arr[] = [1, 1, 1, 2, 2, 2, 3, 3, 3]
              Now 4'th Minimum element is 2

Input  : arr[] = {1, 13, 9, 17, 1, 12}, K = 19, M = 7 
Output : 19'th Minimum element is : 9

简单方法：

1. 将给定的数组附加到向量或称为V的任何其他数组中M次。
2. 按升序对向量或数组V排序。
3. 返回向量V的索引(K-1)处的值，即返回V [K – 1]。

下面是上述方法的实现：

C++

// C++ programme to find the K'th minimum 
// element from an array concatenated M times
  
#include 
using namespace std;
  
// Function to find the K-th minimum element 
// from an array concatenated M times
int KthMinValAfterMconcatenate(int A[], int N, 
                                 int M, int K)
{
    vector V;
  
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++) {
            V.push_back(A[j]);
        }
    }
  
    // sort the elements in ascending order
    sort(V.begin(), V.end());
  
    // return K'th Min element
    // present at K-1 index
    return (V[K - 1]);
}
  
// Driver Code
int main()
{
    int A[] = { 3, 1, 2 };
  
    int M = 3, K = 4;
    int N = sizeof(A) / sizeof(A[0]);
  
    cout << KthMinValAfterMconcatenate(A, N, M, K);
  
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Collections;
  
// Java programme to find the K'th minimum 
// element from an array concatenated M times
class GFG 
{
  
    // Function to find the K-th minimum element 
    // from an array concatenated M times
    static int KthMinValAfterMconcatenate(int[] A, int N,
            int M, int K) 
    {
        ArrayList V = new ArrayList();
  
        for (int i = 0; i < M; i++)
        {
            for (int j = 0; j < N; j++)
            {
                V.add(A[j]);
            }
        }
  
        // sort the elements in ascending order
        Collections.sort(V);
  
        // return K'th Min element
        // present at K-1 index
        return ((int) V.get(K - 1));
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        int[] A = {3, 1, 2};
        int M = 3, K = 4;
        int N = A.length;
        System.out.println(KthMinValAfterMconcatenate(A, N, M, K));
    }
}
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to find the K'th minimum  
# element from an array concatenated M times 
    
# Function to find the K-th minimum element  
# from an array concatenated M times 
def KthMinValAfterMconcatenate(A, N, M, K): 
   
    V = [] 
    
    for i in range(0, M):  
        for j in range(0, N):  
            V.append(A[j]) 
    
    # sort the elements in ascending order 
    V.sort() 
    
    # return K'th Min element 
    # present at K-1 index 
    return V[K - 1] 
    
# Driver Code 
if __name__ == "__main__":
   
    A = [3, 1, 2]  
    
    M, K = 3, 4 
    N = len(A) 
    
    print(KthMinValAfterMconcatenate(A, N, M, K)) 
    
# This code is contributed by Rituraj Jain

C#

// C# programme to find the K'th minimum 
// element from an array concatenated M times
using System;
using System.Collections;
  
class GFG
{
      
// Function to find the K-th minimum element 
// from an array concatenated M times
static int KthMinValAfterMconcatenate(int []A, int N, 
                                int M, int K)
{
    ArrayList V=new ArrayList();
  
    for (int i = 0; i < M; i++) 
    {
        for (int j = 0; j < N; j++)
        {
            V.Add(A[j]);
        }
    }
  
    // sort the elements in ascending order
    V.Sort();
  
    // return K'th Min element
    // present at K-1 index
    return ((int)V[K - 1]);
}
  
// Driver Code
static void Main()
{
    int []A = { 3, 1, 2 };
  
    int M = 3, K = 4;
    int N = A.Length;
  
    Console.WriteLine(KthMinValAfterMconcatenate(A, N, M, K));
}
}
  
// This code is contributed by mits

输出：

2

有效方法：如果M的值较小，则上述方法可以正常工作，但对于
较大的M值将导致内存错误或超时错误。

这样做的目的是要观察到，在对给定数组进行M次连接并再次对其进行排序之后，每个元素现在在新的级联数组中将出现M次。所以，

1. 给定数组升序排序。
2. 返回给定数组的索引((K-1)/ M)上存在的值，即返回arr [((K-1)/ M)]。

下面是上述方法的实现：

C++

// C++ programme to find the K'th minimum element 
// from an array concatenated M times
  
#include 
using namespace std;
  
// Function to find the K-th minimum element 
// from an array concatenated M times
int KthMinValAfterMconcatenate(int A[], int N, 
                                  int M, int K)
{
    // Sort the elements in 
    // ascending order
    sort(A, A + N);
  
    // Return the K'th Min element
    // present at ( (K-1) / M ) index
    return (A[((K - 1) / M)]);
}
  
// Driver Code
int main()
{
    int A[] = { 3, 1, 2 };
  
    int M = 3, K = 4;
    int N = sizeof(A) / sizeof(A[0]);
  
    cout << KthMinValAfterMconcatenate(A, N, M, K);
  
    return 0;
}

Java

// Java programme to find the K'th minimum element 
// from an array concatenated M times
import java.util.*;
  
class GFG1 
{
  
    // Function to find the K-th minimum element 
    // from an array concatenated M times
    static int KthMinValAfterMconcatenate(int[] A, int N,
                                            int M, int K) 
    {
        // Sort the elements in 
        // ascending order
        Arrays.sort(A);
  
        // Return the K'th Min element
        // present at ( (K-1) / M ) index
        return (A[((K - 1) / M)]);
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        int[] A = {3, 1, 2};
  
        int M = 3, K = 4;
        int N = A.length;
  
        System.out.println(KthMinValAfterMconcatenate(A, N, M, K));
    }
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 program to find the K'th minimum 
# element from an array concatenated M times 
  
# Function to find the K-th minimum element 
# from an array concatenated M times 
def KthMinValAfterMconcatenate(A, N, M, K): 
  
    V = [] 
  
    for i in range(0, M): 
        for j in range(0, N): 
            V.append(A[j]) 
  
    # sort the elements in ascending order 
    V.sort() 
  
    # return K'th Min element 
    # present at K-1 index 
    return V[K - 1] 
  
# Driver Code 
if __name__ == "__main__":
  
    A = [3, 1, 2] 
  
    M, K = 3, 4
    N = len(A) 
  
    print(KthMinValAfterMconcatenate(A, N, M, K)) 
  
# This code is contributed by Rituraj Jain

C＃

// C# programme to find the K'th minimum element 
// from an array concatenated M times
using System;
  
class GFG
{
      
// Function to find the K-th minimum element 
// from an array concatenated M times
static int KthMinValAfterMconcatenate(int []A, int N, 
                                int M, int K)
{
    // Sort the elements in 
    // ascending order
    Array.Sort(A);
  
    // Return the K'th Min element
    // present at ( (K-1) / M ) index
    return (A[((K - 1) / M)]);
}
  
// Driver Code
static void Main()
{
    int []A = { 3, 1, 2 };
  
    int M = 3, K = 4;
    int N = A.Length;
  
    Console.WriteLine(KthMinValAfterMconcatenate(A, N, M, K));
}
}
  
// This code is contributed by mits

输出：

2