Find the K-th minimum element from an array concatenated M times
Last Updated :
20 Aug, 2022
Given an array arr[] and two integers K and M. The problem is to find the K-th Minimum element after concatenating the array to itself M times.
Examples:
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
Simple Approach :
- Append the given array into a vector or any other array say V for M times.
- Sort the vector or array V in ascending order.
- Return the value at index ( K-1 ) of vector V i.e., return V[ K - 1 ].
Below is the implementation of the above approach:
C++
// C++ programme to find the K'th minimum
// element from an array concatenated M times
#include <bits/stdc++.h>
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<int> 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;
}
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 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# 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
<?php
// PHP programme 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
function KthMinValAfterMconcatenate($A, $N,
$M, $K)
{
$V = array();
for ($i = 0; $i < $M; $i++)
{
for ($j = 0; $j < $N; $j++)
{
array_push($V, $A[$j]);
}
}
// sort the elements in ascending order
sort($V);
// return K'th Min element
// present at K-1 index
return ($V[$K - 1]);
}
// Driver Code
$A = array( 3, 1, 2 );
$M = 3;
$K = 4;
$N = count($A);
echo KthMinValAfterMconcatenate($A, $N, $M, $K);
// This code is contributed by mits
?>
<script>
// Javascript programme 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
function KthMinValAfterMconcatenate(A, N, M, K)
{
var V = [];
for (var i = 0; i < M; i++) {
for (var j = 0; j < N; j++) {
V.push(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
var A = [3, 1, 2];
var M = 3, K = 4;
var N = A.length;
document.write( KthMinValAfterMconcatenate(A, N, M, K));
</script>
Time Complexity: O((M * N) * log(M * N))
Auxiliary Space: O(M * N)
Efficient Approach: The above approach will work fine if the value of M is small but for
a large value of M will give a memory error or a timeout error.
The idea is to observe that after sorting the given array when we concatenate it M times and sort it again, each element will now appear M times in the new concatenated array. So,
- Sort given array in ascending order.
- Return the value present at the index ( (K-1) / M ) of the given array ie., return arr[((K-1) / M)].
Below is the implementation of the above approach:
C++
// C++ programme to find the K'th minimum element
// from an array concatenated M times
#include <bits/stdc++.h>
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 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 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# 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
<?php
// PHP programme 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
function KthMinValAfterMconcatenate($A, $N, $M, $K)
{
// Sort the elements in
// ascending order
sort($A);
// Return the K'th Min element
// present at ( (K-1) / M ) index
return ($A[(($K - 1) / $M)]);
}
// Driver Code
$A = array(3, 1, 2);
$M = 3; $K = 4;
$N = sizeof($A);
echo(KthMinValAfterMconcatenate($A, $N, $M, $K));
// This code contributed by Code_Mech.
?>
<script>
// JavaScript programme 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
function KthMinValAfterMconcatenate(A, N, M, K)
{
// Sort the elements in
// ascending order
A.sort((a,b)=>a-b)
// Return the K'th Min element
// present at ( (K-1) / M ) index
return (A[((K - 1) / M)]);
}
// Driver Code
var A = [3, 1, 2 ];
var M = 3, K = 4;
var N = A.length;
document.write( KthMinValAfterMconcatenate(A, N, M, K));
</script>
Time Complexity: O(N * log N)
Auxiliary Space: O(1), since no extra space has been taken.
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