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

Difficulty Level : Medium
Last Updated : 25 Aug, 2021

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 all array elements are distinct.

Examples:

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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++ program of above implementation
#include<bits/stdc++.h>
using namespace std;
 
// 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
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() % 100 + 1) % n;
    swap(arr[l + pivot], arr[r]);
    return partition(arr, l, 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 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_MAX;
}
 
// Driver Code
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);
}
 
// This code is contributed by Surbhi Tyagi.

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(n);
        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

Javascript

<script>
 
// Javascript program 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
function  kthSmallest(arr, l, r, 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
        let 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 Number.MAX_VALUE;
}
 
function swap(arr, i, j)
{
    let 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
function partition(arr, l, r)
{
    let x = arr[r], i = l;
    for(let j = l; j <= r - 1; j++)
    {
        if (arr[j] <= x)
        {
            swap(arr, i, j);
            i++;
        }
    }
    swap(arr, i, r);
    return i;
}
 
function randomPartition(arr, l, r)
{
    let n = r - l + 1;
    let pivot = (Math.floor(Math.random() * 101)) % n;
    swap(arr, l + pivot, r);
     
    return partition(arr, l, r);
}
 
// Driver code
let arr = [ 12, 3, 5, 7, 4, 19, 26 ];
let n = arr.length, k = 3;
 
document.write("K'th smallest element is " +
               kthSmallest(arr, 0, n - 1, k));
 
// This code is contributed by unknown2108
     
</script>

Output

K'th smallest element is 5

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

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

C++

Java

Python3

C#

Javascript

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