K'th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)

We recommend to read following post as a prerequisite of this post.

K’th Smallest/Largest Element in Unsorted Array | Set 1

Given an array and a number k where k is smaller than size of array, we need to find the k’th largest element in the given array. It is given that ll array elements are distinct.

Examples:

Input: arr[] = {7, 10, 4, 3, 20, 15}
       k = 3
Output: 7

Input: arr[] = {7, 10, 4, 3, 20, 15}
       k = 4
Output: 10

We have discussed three different solutions here.

In this post method 4 is discussed which is mainly an extension of method 3 (QuickSelect) discussed in the previous post.

C++

// C++ implementation of above implementation 
#include<iostream> 
#include<climits> 
#include<cstdlib> 
using namespace std; 
  
int randomPartition(int arr[], int l, int r); 
  
// This function returns k'th smallest element in arr[l..r] using 
// QuickSort based method.  ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT 
int kthSmallest(int arr[], int l, int r, int k) 
{ 
    // If k is smaller than number of elements in array 
    if (k > 0 && k <= r - l + 1) 
    { 
        // Partition the array around last element and get 
        // position of pivot element in sorted array 
        int pos = randomPartition(arr, l, r); 
  
        // If position is same as k 
        if (pos-l == k-1) 
            return arr[pos]; 
        if (pos-l > k-1)  // If position is more, recur for left subarray 
            return kthSmallest(arr, l, pos-1, k); 
  
        // Else recur for right subarray 
        return kthSmallest(arr, pos+1, r, k-pos+l-1); 
    } 
  
    // If k is more than number of elements in array 
    return INT_MAX; 
} 
  
void swap(int *a, int *b) 
{ 
    int temp = *a; 
    *a = *b; 
    *b = temp; 
} 
  
// Standard partition process of QuickSort().  It considers the last 
// element as pivot and moves all smaller element to left of it 
// and greater elements to right 
int partition(int arr[], int l, int r) 
{ 
    int x = arr[r], i = l; 
    for (int j = l; j <= r - 1; j++) 
    { 
        if (arr[j] <= x) 
        { 
            swap(&arr[i], &arr[j]); 
            i++; 
        } 
    } 
    swap(&arr[i], &arr[r]); 
    return i; 
} 
  
int randomPartition(int arr[], int l, int r) 
{ 
    int n = r-l+1; 
    int pivot = rand() % n; 
    swap(&arr[l + pivot], &arr[r]); 
    return partition(arr, l, r); 
} 
  
// Driver program to test above methods 
int main() 
{ 
    int arr[] = {12, 3, 5, 7, 4, 19, 26}; 
    int n = sizeof(arr)/sizeof(arr[0]), k = 3; 
    cout << "K'th smallest element is " << kthSmallest(arr, 0, n-1, k); 
    return 0; 
}                   

Java

// Java program of above implementation 
import java.util.Random; 
  
public class GFG { 
  
// This function returns k'th smallest element in arr[l..r] using 
// QuickSort based method. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT 
    static int kthSmallest(int arr[], int l, int r, int k) { 
        // If k is smaller than number of elements in array 
        if (k > 0 && k <= r - l + 1) { 
            // Partition the array around last element and get 
            // position of pivot element in sorted array 
            int pos = randomPartition(arr, l, r); 
  
            // If position is same as k 
            if (pos - l == k - 1) { 
                return arr[pos]; 
            } 
            if (pos - l > k - 1) // If position is more, recur for left subarray 
            { 
                return kthSmallest(arr, l, pos - 1, k); 
            } 
  
            // Else recur for right subarray 
            return kthSmallest(arr, pos + 1, r, k - pos + l - 1); 
        } 
  
        // If k is more than number of elements in array 
        return Integer.MAX_VALUE; 
    } 
  
    static void swap(int[] arr, int i, int j) { 
        int temp = arr[i]; 
        arr[i] = arr[j]; 
        arr[j] = temp; 
    } 
  
// Standard partition process of QuickSort(). It considers the last 
// element as pivot and moves all smaller element to left of it 
// and greater elements to right 
    static int partition(int arr[], int l, int r) { 
        int x = arr[r], i = l; 
        for (int j = l; j <= r - 1; j++) { 
            if (arr[j] <= x) { 
                swap(arr, i, j); 
                i++; 
            } 
        } 
        swap(arr, i, r); 
        return i; 
    } 
  
    static int randomPartition(int arr[], int l, int r) { 
        int n = r - l + 1; 
        int pivot = new Random().nextInt(1); 
        swap(arr, l + pivot, r); 
        return partition(arr, l, r); 
    } 
  
// Driver program to test above methods 
    public static void main(String args[]) { 
        int arr[] = {12, 3, 5, 7, 4, 19, 26}; 
        int n = arr.length, k = 3; 
        System.out.println("K'th smallest element is " + kthSmallest(arr, 0, n - 1, k)); 
    } 
} 
  
/*This code is contributed by 29AjayKumar*/

Python3

# Python3 implementation of above implementation 
  
# This function returns k'th smallest element  
# in arr[l..r] using QuickSort based method. 
# ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT 
from random import randint 
  
def randomPartition(arr, l, r): 
    n = r - l + 1
    pivot = randint(1, 100) % n 
    arr[l + pivot], arr[r] = arr[l + pivot], arr[r] 
    return partition(arr, l, r) 
  
def kthSmallest(arr, l, r, k): 
  
    # If k is smaller than  
    # number of elements in array 
    if (k > 0 and k <= r - l + 1): 
  
        # Partition the array around last element and  
        # get position of pivot element in sorted array 
        pos = randomPartition(arr, l, r) 
  
        # If position is same as k 
        if (pos - l == k - 1): 
            return arr[pos] 
              
        # If position is more, recur for left subarray 
        if (pos - l > k - 1):  
            return kthSmallest(arr, l, pos - 1, k) 
  
        # Else recur for right subarray 
        return kthSmallest(arr, pos + 1, r, 
                           k - pos + l - 1) 
  
    # If k is more than number of elements in array 
    return 10**9
  
# Standard partition process of QuickSort().  
# It considers the last element as pivot and  
# moves all smaller element to left of it 
# and greater elements to right 
def partition(arr, l, r): 
    x = arr[r] 
    i = l 
    for j in range(l, r): 
        if (arr[j] <= x): 
            arr[i], arr[j] = arr[j], arr[i] 
            i += 1
  
    arr[i], arr[r] = arr[r], arr[i] 
    return i 
  
# Driver Code 
arr = [12, 3, 5, 7, 4, 19, 26] 
n = len(arr) 
k = 3
print("K'th smallest element is",  
       kthSmallest(arr, 0, n - 1, k)) 
  
# This code is contributed by Mohit Kumar 

C#

// C# program of above implementation 
using System; 
  
class GFG  
{  
  
// This function returns k'th smallest  
// element in arr[l..r] using  
// QuickSort based method. ASSUMPTION:  
// ALL ELEMENTS IN ARR[] ARE DISTINCT  
static int kthSmallest(int []arr, int l, 
                       int r, int k)  
{  
    // If k is smaller than number  
    // of elements in array  
    if (k > 0 && k <= r - l + 1)  
    { 
        // Partition the array around last  
        // element and get position of pivot 
        // element in sorted array  
        int pos = randomPartition(arr, l, r);  
  
        // If position is same as k  
        if (pos - l == k - 1)  
        {  
            return arr[pos];  
        }  
          
        // If position is more, recur  
        // for left subarray  
        if (pos - l > k - 1)  
        {  
            return kthSmallest(arr, l, pos - 1, k);  
        }  
  
        // Else recur for right subarray  
        return kthSmallest(arr, pos + 1, r, 
                           k - pos + l - 1);  
    }  
  
    // If k is more than number of  
    // elements in array  
    return int.MaxValue;  
}  
  
static void swap(int[] arr, int i, int j) 
{  
    int temp = arr[i];  
    arr[i] = arr[j];  
    arr[j] = temp;  
}  
  
// Standard partition process of QuickSort().  
// It considers the last element as pivot and  
// oves all smaller element to left of it  
// and greater elements to right  
static int partition(int []arr, int l, int r)  
{  
    int x = arr[r], i = l;  
    for (int j = l; j <= r - 1; j++) 
    {  
        if (arr[j] <= x)  
        {  
            swap(arr, i, j);  
            i++;  
        }  
    }  
    swap(arr, i, r);  
    return i;  
}  
  
static int randomPartition(int []arr, int l, int r) 
{  
    int n = r - l + 1;  
    int pivot = new Random().Next(1);  
    swap(arr, l + pivot, r);  
    return partition(arr, l, r);  
}  
  
// Driver Code 
public static void Main() 
{  
    int []arr = {12, 3, 5, 7, 4, 19, 26};  
    int n = arr.Length, k = 3;  
    Console.WriteLine("K'th smallest element is " +  
                    kthSmallest(arr, 0, n - 1, k));  
}  
}  
  
// his code is contributed by 29AjayKumar 

Output:

K'th smallest element is 5

References:
https://www.geeksforgeeks.org/kth-smallestlargest-element-unsorted-array/

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

My Personal Notes

K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)

C++

Java

Python3

C#

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