Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Quickselect Algorithm

  • Difficulty Level : Medium
  • Last Updated : 10 Aug, 2021

Quickselect is a selection algorithm to find the k-th smallest element in an unordered list. It is related to the quick sort sorting algorithm.
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

The algorithm is similar to QuickSort. The difference is, instead of recurring for both sides (after finding pivot), it recurs only for the part that contains the k-th smallest element. The logic is simple, if index of partitioned element is more than k, then we recur for left part. If index is same as k, we have found the k-th smallest element and we return. If index is less than k, then we recur for right part. This reduces the expected complexity from O(n log n) to O(n), with a worst case of O(n^2).

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

function quickSelect(list, left, right, k)

   if left = right
      return list[left]

   Select a pivotIndex between left and right

   pivotIndex := partition(list, left, right, 
                                  pivotIndex)
   if k = pivotIndex
      return list[k]
   else if k < pivotIndex
      right := pivotIndex - 1
   else
      left := pivotIndex + 1 

C++14




// CPP program for implementation of QuickSelect
#include <bits/stdc++.h>
using namespace std;
 
// 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;
}
 
// 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 index = partition(arr, l, r);
 
        // If position is same as k
        if (index - l == k - 1)
            return arr[index];
 
        // If position is more, recur
        // for left subarray
        if (index - l > k - 1)
            return kthSmallest(arr, l, index - 1, k);
 
        // Else recur for right subarray
        return kthSmallest(arr, index + 1, r,
                            k - index + l - 1);
    }
 
    // If k is more than number of
    // elements in array
    return INT_MAX;
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 10, 4, 5, 8, 6, 11, 26 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << "K-th smallest element is "
        << kthSmallest(arr, 0, n - 1, k);
    return 0;
}

Java




// Java program of Quick Select
import java.util.Arrays;
 
class GFG {
 
    // partition function similar to quick sort
    // Considers last element as pivot and adds
    // elements with less value to the left and
    // high value to the right and also changes
    // the pivot position to its respective position
    // in the final array.
    public static int partition(int[] arr, int low,
                                int high)
    {
        int pivot = arr[high], pivotloc = low;
        for (int i = low; i <= high; i++) {
            // inserting elements of less value
            // to the left of the pivot location
            if (arr[i] < pivot) {
                int temp = arr[i];
                arr[i] = arr[pivotloc];
                arr[pivotloc] = temp;
                pivotloc++;
            }
        }
 
        // swapping pivot to the final pivot location
        int temp = arr[high];
        arr[high] = arr[pivotloc];
        arr[pivotloc] = temp;
 
        return pivotloc;
    }
 
    // finds the kth position (of the sorted array)
    // in a given unsorted array i.e this function
    // can be used to find both kth largest and
    // kth smallest element in the array.
    // ASSUMPTION: all elements in arr[] are distinct
    public static int kthSmallest(int[] arr, int low,
                                  int high, int k)
    {
        // find the partition
        int partition = partition(arr, low, high);
 
        // if partition value is equal to the kth position,
        // return value at k.
        if (partition == k - 1)
            return arr[partition];
 
        // if partition value is less than kth position,
        // search right side of the array.
        else if (partition < k - 1)
            return kthSmallest(arr, partition + 1, high, k);
 
        // if partition value is more than kth position,
        // search left side of the array.
        else
            return kthSmallest(arr, low, partition - 1, k);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] array = new int[] { 10, 4, 5, 8, 6, 11, 26 };
        int[] arraycopy
            = new int[] { 10, 4, 5, 8, 6, 11, 26 };
 
        int kPosition = 3;
        int length = array.length;
 
        if (kPosition > length) {
            System.out.println("Index out of bound");
        }
        else {
            // find kth smallest value
            System.out.println(
                "K-th smallest element in array : "
                + kthSmallest(arraycopy, 0, length - 1,
                              kPosition));
        }
    }
}
 
// This code is contributed by Saiteja Pamulapati

Python3




# Python3 program of Quick Select
 
# 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
 
# finds the kth position (of the sorted array)
# in a given unsorted array i.e this function
# can be used to find both kth largest and
# kth smallest element in the array.
# ASSUMPTION: all 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 last
        # element and get position of pivot
        # element in sorted array
        index = partition(arr, l, r)
 
        # if position is same as k
        if (index - l == k - 1):
            return arr[index]
 
        # If position is more, recur
        # for left subarray
        if (index - l > k - 1):
            return kthSmallest(arr, l, index - 1, k)
 
        # Else recur for right subarray
        return kthSmallest(arr, index + 1, r,
                            k - index + l - 1)
    print("Index out of bound")
 
# Driver Code
arr = [ 10, 4, 5, 8, 6, 11, 26 ]
n = len(arr)
k = 3
print("K-th smallest element is ", end = "")
print(kthSmallest(arr, 0, n - 1, k))
 
# This code is contributed by Muskan Kalra.

C#




// C# program of Quick Select
using System;
 
class GFG
{
     
    // partition function similar to quick sort
    // Considers last element as pivot and adds
    // elements with less value to the left and
    // high value to the right and also changes
    // the pivot position to its respective position
    // in the readonly array.
    static int partitions(int []arr,int low, int high)
    {
        int pivot = arr[high], pivotloc = low, temp;
        for (int i = low; i <= high; i++)
        {
            // inserting elements of less value
            // to the left of the pivot location
            if(arr[i] < pivot)
            {        
                temp = arr[i];
                arr[i] = arr[pivotloc];
                arr[pivotloc] = temp;
                pivotloc++;
            }
        }
         
        // swapping pivot to the readonly pivot location
        temp = arr[high];
        arr[high] = arr[pivotloc];
        arr[pivotloc] = temp;
         
        return pivotloc;
    }
     
    // finds the kth position (of the sorted array)
    // in a given unsorted array i.e this function
    // can be used to find both kth largest and
    // kth smallest element in the array.
    // ASSUMPTION: all elements in []arr are distinct
    static int kthSmallest(int[] arr, int low,
                                int high, int k)
    {
        // find the partition
        int partition = partitions(arr,low,high);
 
        // if partition value is equal to the kth position,
        // return value at k.
        if(partition == k)
            return arr[partition];
             
        // if partition value is less than kth position,
        // search right side of the array.
        else if(partition < k )
            return kthSmallest(arr, partition + 1, high, k );
             
        // if partition value is more than kth position,
        // search left side of the array.
        else
            return kthSmallest(arr, low, partition - 1, k );        
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        int[] array = {10, 4, 5, 8, 6, 11, 26};
        int[] arraycopy = {10, 4, 5, 8, 6, 11, 26};
                 
        int kPosition = 3;
        int length = array.Length;
         
        if(kPosition > length)
        {
            Console.WriteLine("Index out of bound");
        }
        else
        {
            // find kth smallest value
            Console.WriteLine("K-th smallest element in array : " +
                                kthSmallest(arraycopy, 0, length - 1,
                                                        kPosition - 1));
        }
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript program of Quick Select
 
// partition function similar to quick sort
    // Considers last element as pivot and adds
    // elements with less value to the left and
    // high value to the right and also changes
    // the pivot position to its respective position
    // in the final array.
function _partition(arr, low, high)
{
    let pivot = arr[high], pivotloc = low;
        for (let i = low; i <= high; i++)
        {
         
            // inserting elements of less value
            // to the left of the pivot location
            if (arr[i] < pivot)
            {
                let temp = arr[i];
                arr[i] = arr[pivotloc];
                arr[pivotloc] = temp;
                pivotloc++;
            }
        }
  
        // swapping pivot to the final pivot location
        let temp = arr[high];
        arr[high] = arr[pivotloc];
        arr[pivotloc] = temp;
  
        return pivotloc;
}
 
// finds the kth position (of the sorted array)
    // in a given unsorted array i.e this function
    // can be used to find both kth largest and
    // kth smallest element in the array.
    // ASSUMPTION: all elements in arr[] are distinct
function kthSmallest(arr, low, high, k)
{
 
    // find the partition
        let partition = _partition(arr, low, high);
  
        // if partition value is equal to the kth position,
        // return value at k.
        if (partition == k - 1)
            return arr[partition];
  
        // if partition value is less than kth position,
        // search right side of the array.
        else if (partition < k - 1)
            return kthSmallest(arr, partition + 1, high, k);
  
        // if partition value is more than kth position,
        // search left side of the array.
        else
            return kthSmallest(arr, low, partition - 1, k);
}
 
// Driver Code
let array = [ 10, 4, 5, 8, 6, 11, 26];
let arraycopy = [10, 4, 5, 8, 6, 11, 26 ];
let kPosition = 3;
let length = array.length;
 
if (kPosition > length) {
    document.write("Index out of bound<br>");
}
else
{
 
// find kth smallest value
document.write(
"K-th smallest element in array : "
+ kthSmallest(arraycopy, 0, length - 1,
              kPosition)+"<br>");
}
 
// This code is contributed by rag2127
</script>

Output: 

K-th smallest element is 6

Important Points:  

  1. Like quicksort, it is fast in practice, but has poor worst-case performance. It is used in
  2. The partition process is same as QuickSort, only recursive code differs.
  3. There exists an algorithm that finds k-th smallest element in O(n) in worst case, but QuickSelect performs better on average. 
     

Related C++ function : std::nth_element in C++
This article is contributed by Sahil Chhabra . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 




My Personal Notes arrow_drop_up
Recommended Articles
Page :