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

We recommend reading the 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 the size of the array, we need to find the k’th smallest element in the given array. It is given that all 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. The idea is to randomly pick a pivot element. To implement randomized partition, we use a random function, rand() to generate index between l and r, swap the element at randomly generated index with the last element, and finally call the standard partition process which uses last element as pivot.

Following is an implementation of the above Randomized QuickSelect.

C/C++

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// C++ implementation of randomized quickSelect
#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: 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 a random 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 the number of elements in the 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. This function is used by randomPartition()
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;
}
  
// Picks a random pivot element between l and r and partitions
// arr[l..r] around the randomly picked element using partition()
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;
}

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Java

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// Java program to find k'th smallest element in expected
// linear time
class KthSmallst
{
    // 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 a random 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 Integer.MAX_VALUE;
    }
  
    // Utility method to swap arr[i] and arr[j]
    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. This function
    // is used by randomPartition()
    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;
    }
  
    // Picks a random pivot element between l and r and 
    // partitions arr[l..r] arount the randomly picked 
    // element using partition()
    int randomPartition(int arr[], int l, int r)
    {
        int n = r-l+1;
        int pivot = (int)(Math.random()) * (n-1);
        swap(arr, l + pivot, r);
        return partition(arr, l, r);
    }
  
    // Driver method to test above
    public static void main(String args[])
    {
        KthSmallst ob = new KthSmallst();
        int arr[] = {12, 3, 5, 7, 4, 19, 26};
        int n = arr.length,k = 3;
        System.out.println("K'th smallest element is "+
                           ob.kthSmallest(arr, 0, n-1, k));
    }
}
/*This code is contributed by Rajat Mishra*/

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Python3

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# Python3 implementation of randomized 
# quickSelect 
import random
  
# This function returns k'th smallest 
# element in arr[l..r] using QuickSort
# based method. ASSUMPTION: ELEMENTS
# IN ARR[] ARE DISTINCT 
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 a random 
        # 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 (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 the number of 
    # elements in the array 
    return 999999999999
  
def swap(arr, a, b):
    temp = arr[a]
    arr[a] = arr[b]
    arr[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. This function
# is used by randomPartition() 
def partition(arr, l, r):
    x = arr[r]
    i = l
    for j in range(l, r):
        if (arr[j] <= x):
            swap(arr, i, j) 
            i += 1
    swap(arr, i, r) 
    return i
  
# Picks a random pivot element between l and r 
# and partitions arr[l..r] around the randomly
# picked element using partition() 
def randomPartition(arr, l, r):
    n = r - l + 1
    pivot = int(random.random() % n) 
    swap(arr, l + pivot, r) 
    return partition(arr, l, r)
  
# Driver Code
if __name__ == '__main__':
  
    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 PranchalK

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C#

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// C# program to find k'th smallest 
// element in expected linear time 
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 
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 a 
        // random 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; 
  
// Utility method to swap arr[i] and arr[j] 
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. This function 
// is used by randomPartition() 
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; 
  
// Picks a random pivot element between 
// l and r and partitions arr[l..r] 
// around the randomly picked element
// using partition() 
int randomPartition(int []arr, int l, int r) 
    int n = r - l + 1; 
    Random rnd = new Random();
    int rand = rnd.Next(0, 1);
    int pivot = (int)(rand * (n - 1)); 
    swap(arr, l + pivot, r); 
    return partition(arr, l, r); 
  
// Driver Code
public static void Main() 
    GFG ob = new GFG(); 
    int []arr = {12, 3, 5, 7, 4, 19, 26}; 
    int n = arr.Length,k = 3; 
    Console.Write("K'th smallest element is "
            ob.kthSmallest(arr, 0, n - 1, k)); 
  
// This code is contributed by 29AjayKumar

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PHP

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<?php
// C# program to find k'th smallest 
// element in expected linear time 
  
// This function returns k'th smallest  
// element in arr[l..r] using QuickSort based method.
// ASSUMPTION: 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 a random 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 the number of elements in the array 
    return PHP_INT_MAX; 
  
function swap($a, $b
    $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.
// This function is used by randomPartition() 
function partition($arr, $l, $r
    $x = $arr[$r];
    $i = $l
    for ($j = $l; $j <= $r - 1; $j++) 
    
        if ($arr[$j] <= $x
        
            list($arr[$i], $arr[$j])=array($arr[$j],$arr[$i]);
            //swap(&arr[i], &arr[j]); 
            $i++; 
        
    
    list($arr[$i], $arr[$r])=array($arr[$r],$arr[$i]);
    //swap(&arr[i], &arr[r]); 
    return $i
  
// Picks a random pivot element between
//  l and r and partitions arr[l..r] around 
// the randomly picked element using partition() 
function randomPartition($arr, $l, $r
    $n = $r-$l+1; 
    $pivot = rand() % $n
      
    list($arr[$l + $pivot], $arr[$r]) = 
            array($arr[$r],$arr[$l + $pivot] );
      
    //swap(&arr[l + pivot], &arr[r]); 
    return partition($arr, $l, $r); 
  
// Driver program to test the above methods 
    $arr = array(12, 3, 5, 7, 4, 19, 260); 
    $n = sizeof($arr)/sizeof($arr[0]);
    $k = 3; 
    echo "K'th smallest element is " ,
            kthSmallest($arr, 0, $n-1, $k); 
      
  
// This code is contributed by ajit.
?>

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Output:

K'th smallest element is 5 

Time Complexity:
The worst case time complexity of the above solution is still O(n2). In worst case, the randomized function may always pick a corner element. The expected time complexity of above randomized QuickSelect is O(n), see CLRS book or MIT video lecture for proof. The assumption in the analysis is, random number generator is equally likely to generate any number in the input range.

Sources:
MIT Video Lecture on Order Statistics, Median
Introduction to Algorithms by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L.

This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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