Floyd-Rivest Algorithm

The Floyd-Rivest algorithm is a selection algorithm used to find the kth smallest element in an array of distinct elements. It is similar to the QuickSelect algorithm but has better running time in practice.
Like QuickSelect the algorithm works on the idea of partitioning. After partitioning an array, the partition element ends up in the corrected sorted position. If the array has all distinct elements, retrieving the (k+1)th smallest element is same as retrieving the (k+1)th element after sorting. Because a full sort is expensive(takes O(N log N) to compute), the Floyd-Rivest algorithm leverages partitioning to accomplish the same in O(N) time.

Algorithm:

  1. If the size of the array S being considered is small enough, then the QuickSelect algorithm is applied directly to get the K-th smallest element. This size is an arbitrary constant of the algorithm, which the authors chose as 600.
  2. Otherwise, 2 pivots are chosen- newLeftIndex and newRightIndex using random sampling such that they have the highest probability of containing the K-th largest element. Then, the function is called recursively with left and right boundaries of the array now set to newLeftIndex and newRightIndex.
  3. Like QuickSelect, after partitioning the subarray, the left and right boundaries need to be set such that they contain the K-largest element.
    After partitioning the array around K, the Kth element is in its correct sorted position. So, the left and right boundaries are set in such a way that the subarray being considered contains array[k]

Below is the implementation of the above approach.

C++



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// C++ implementation of the above approach.
#include <iostream>
#include <math.h>
using namespace std;
  
// Function to return the
// sign of a number
int sign(double x)
{
    if (x < 0)
        return -1;
    if (x > 0)
        return 1;
    return 0;
}
  
// Function to swap
// two numbers in an array.
void swap(int arr[], int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}
  
int select(int arr[], int left,
           int right, int k)
{
    while (right > left) {
        if (right - left > 600) {
            // Choosing a small subarray
            // S based on sampling.
            // 600, 0.5 and 0.5
            // are arbitrary constants
            int n = right - left + 1;
            int i = k - left + 1;
            double z = log(n);
            double s = 0.5 * exp(2 * z / 3);
            double sd = 0.5 * sqrt(z * s
                                   * (n - s) / n)
                        * sign(i - n / 2);
  
            int newLeft = max(left,
                              (int)(k - i * s / n + sd));
  
            int newRight = min(right,
                               (int)(k + (n - i) * s / n
                                     + sd));
  
            select(arr, newLeft, newRight, k);
        }
  
        // Partition the subarray S[left..right]
        // with arr[k] as pivot
        int t = arr[k];
        int i = left;
        int j = right;
        swap(arr, left, k);
        if (arr[right] > t) {
            swap(arr, left, right);
        }
  
        while (i < j) {
            swap(arr, i, j);
            i++;
            j--;
  
            while (arr[i] < t)
                i++;
            while (arr[j] > t)
                j--;
        }
  
        if (arr[left] == t)
            swap(arr, left, j);
        else {
            j++;
            swap(arr, right, j);
        }
  
        // Adjust the left and right pointers
        // to select the subarray having k
        if (j <= k)
            left = j + 1;
        if (k <= j)
            right = j - 1;
    }
    return arr[k];
}
  
// Driver code
int main()
{
    int arr[] = { 7, 3, 4, 0, 1, 6 };
    int n = sizeof(arr) / sizeof(int);
  
    // k-th smallest element.
    // In this we get the 2nd smallest element
    int k = 2;
    int res = select(arr, 0, n - 1, k - 1);
    cout << "The " << k << "-th smallest element is "
         << res << endl;
    return 0;
}

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Java

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// Java implementation of the above approach.
class GFG {
  
    // Function to return
    // the sign of the number
    int sign(double x)
    {
        if (x < 0)
            return -1;
        if (x > 0)
            return 1;
        return 0;
    }
  
    // Function to swap two numbers in an array
    void swap(int arr[], int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }
  
    // Function to return kth smallest number
    int select(int arr[], int left,
               int right, int k)
    {
        while (right > left) {
            if (right - left > 600) {
                // Choosing a small subarray
                // S based on sampling.
                // 600, 0.5 and 0.5 are
                // arbitrary constants
                int n = right - left + 1;
                int i = k - left + 1;
                double z = Math.log(n);
                double s = 0.5 * Math.exp(2 * z / 3);
  
                double sd = 0.5 * Math.sqrt(z * s * (n - s) / n)
                            * sign(i - n / 2);
  
                int newLeft = Math.max(left,
                                       (int)(k - i * s / n
                                             + sd));
  
                int newRight = Math.min(right,
                                        (int)(k + (n - i) * s / n
                                              + sd));
  
                select(arr, newLeft, newRight, k);
            }
  
            // Partition the subarray S[left..right]
            // with arr[k] as pivot
            int t = arr[k];
            int i = left;
            int j = right;
            swap(arr, left, k);
            if (arr[right] > t) {
                swap(arr, left, right);
            }
  
            while (i < j) {
                swap(arr, i, j);
                i++;
                j--;
  
                while (arr[i] < t)
                    i++;
                while (arr[j] > t)
                    j--;
            }
  
            if (arr[left] == t)
                swap(arr, left, j);
            else {
                j++;
                swap(arr, right, j);
            }
  
            // Adjust the left and right
            // pointers to select the subarray having k
            if (j <= k)
                left = j + 1;
            if (k <= j)
                right = j - 1;
        }
        return arr[k];
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = new int[] { 7, 3, 4, 0, 1, 6 };
  
        // k-th smallest element.
        // In this we get the 2nd smallest element
        int k = 2;
        FloydRivest f = new FloydRivest();
        int res = f.select(arr, 0, arr.length - 1, k - 1);
        System.out.println("The " + k
                           + "-th smallest element is " + res);
    }
}

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Python3

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# Python implementation of the above approach.
import math
import random
  
# Function to return the 
# sign of the number
def sign(x):
    if x>0:
        return 1
    elif x<0:
        return -1
    return 0
  
# Function to swap two 
# numbers in an array
def swap(arr, i, j):
    temp = arr[i]
    arr[i] = arr[j]
    arr[j] = temp
  
# Function to return kth smallest number
def select(arr: list, left: int
right: int, k: int):
    while right>left:
  
        # Choosing a small subarray
        # S based on sampling.
        # 600, 0.5 and 0.5 are
        # arbitrary constants
        if right-left > 600:
            n = right - left + 1
            i = k - left + 1
            z = math.log(n)
            s = 0.5 * math.exp(2 * z / 3)
            sd = 0.5 * math.sqrt(z * s * (n-s)/n) * sign(i-n / 2)
            newLeft = int(max(left, k-i * s / n + sd))
            newRight = int(min(right, k + (n - i) * s / n + sd))
            select(arr, newLeft, newRight, k)
        t = arr[k]
        i = left
        j = right
        swap(arr, left, k)
        if arr[right] > t:
            swap(arr, left, right)
        while i<j:
            swap(arr, i, j)
            i = i + 1
            j = j-1
            while arr[i]<t:
                i = i + 1
            while arr[j] >t:
                j = j-1
  
        if arr[left] == t:
            swap(arr, left, j)
        else:
            j = j + 1
            swap(arr, right, j)
  
        # Updating the left and right indices 
        # depending on position of k-th element 
        if j<= k:
            left = j + 1
        if k<= j:
            right = j-1
    return arr[k]
  
  
arr = [7, 3, 4, 0, 1, 6]
# k-th smallest element. 
# In this the 2nd smallest element is returned.
k = 2
res = select(arr, 0, len(arr)-1, k-1)
print('The {}-th smallest element is {}'.format(k, res)) 

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C#

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// C# implementation of the above approach. 
using System;
  
class GFG 
  
    // Function to return 
    // the sign of the number 
    static int sign(double x) 
    
        if (x < 0) 
            return -1; 
        if (x > 0) 
            return 1; 
        return 0; 
    
  
    // Function to swap two numbers in an array 
    static void swap(int []arr, int i, int j) 
    
        int temp = arr[i]; 
        arr[i] = arr[j]; 
        arr[j] = temp; 
    
  
    // Function to return kth smallest number 
    static int select(int []arr, int left, 
            int right, int k) 
    
        int i;
        while (right > left) 
        
            if (right - left > 600)
            
                // Choosing a small subarray 
                // S based on sampling. 
                // 600, 0.5 and 0.5 are 
                // arbitrary constants 
                int n = right - left + 1; 
                i = k - left + 1; 
                double z = Math.Log(n); 
                double s = 0.5 * Math.Exp(2 * z / 3); 
  
                double sd = 0.5 * Math.Sqrt(z * s * (n - s) / n) 
                            * sign(i - n / 2); 
  
                int newLeft = Math.Max(left, 
                                    (int)(k - i * s / n 
                                            + sd)); 
  
                int newRight = Math.Min(right, 
                                        (int)(k + (n - i) * s / n 
                                            + sd)); 
  
                select(arr, newLeft, newRight, k); 
            
  
            // Partition the subarray S[left..right] 
            // with arr[k] as pivot 
            int t = arr[k]; 
            i = left; 
            int j = right; 
            swap(arr, left, k); 
            if (arr[right] > t)
            
                swap(arr, left, right); 
            
  
            while (i < j)
            
                swap(arr, i, j); 
                i++; 
                j--; 
  
                while (arr[i] < t) 
                    i++; 
                while (arr[j] > t) 
                    j--; 
            
  
            if (arr[left] == t) 
                swap(arr, left, j); 
            else 
            
                j++; 
                swap(arr, right, j); 
            
  
            // Adjust the left and right 
            // pointers to select the subarray having k 
            if (j <= k) 
                left = j + 1; 
            if (k <= j) 
                right = j - 1; 
        
        return arr[k]; 
    
  
    // Driver code 
    public static void Main() 
    
        int[] arr = { 7, 3, 4, 0, 1, 6 }; 
  
        // k-th smallest element. 
        // In this we get the 2nd smallest element 
        int k = 2; 
          
        int res = select(arr, 0, arr.Length - 1, k - 1); 
        Console.WriteLine("The " + k + "-th smallest element is " + res); 
    
  
// This code is contributed by AnkitRai01

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Output:

The 2-th smallest element is 1

Time complexity: O(N)

Reference:
Wikipedia-Floyd Rivest
On Floyd and Rivest’s SELECT algorithm
Wikipedia-Quickselect

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