Shuffle a given array using Fisher–Yates shuffle Algorithm

Given an array, write a program to generate a random permutation of array elements. This question is also asked as “shuffle a deck of cards” or “randomize a given array”. Here shuffle means that every permutation of array element should equally likely.

shuffle-array
Let the given array be arr[]. A simple solution is to create an auxiliary array temp[] which is initially a copy of arr[]. Randomly select an element from temp[], copy the randomly selected element to arr[0] and remove the selected element from temp[]. Repeat the same process n times and keep copying elements to arr[1], arr[2], … . The time complexity of this solution will be O(n^2).

Fisher–Yates shuffle Algorithm works in O(n) time complexity. The assumption here is, we are given a function rand() that generates random number in O(1) time.
The idea is to start from the last element, swap it with a randomly selected element from the whole array (including last). Now consider the array from 0 to n-2 (size reduced by 1), and repeat the process till we hit the first element.

Following is the detailed algorithm

To shuffle an array a of n elements (indices 0..n-1):
  for i from n - 1 downto 1 do
       j = random integer with 0 <= j <= i
       exchange a[j] and a[i]

Following is implementation of this algorithm.

C++

// C++ Program to shuffle a given array  
#include<bits/stdc++.h>  
#include <stdlib.h>  
#include <time.h>  
using namespace std;  
  
// A utility function to swap to integers  
void swap (int *a, int *b)  
{  
    int temp = *a;  
    *a = *b;  
    *b = temp;  
}  
  
// A utility function to print an array  
void printArray (int arr[], int n)  
{  
    for (int i = 0; i < n; i++)  
        cout << arr[i] << " ";  
    cout << "\n";  
}  
  
// A function to generate a random  
// permutation of arr[]  
void randomize (int arr[], int n)  
{  
    // Use a different seed value so that  
    // we don't get same result each time 
    // we run this program  
    srand (time(NULL));  
  
    // Start from the last element and swap  
    // one by one. We don't need to run for  
    // the first element that's why i > 0  
    for (int i = n - 1; i > 0; i--)  
    {  
        // Pick a random index from 0 to i  
        int j = rand() % (i + 1);  
  
        // Swap arr[i] with the element  
        // at random index  
        swap(&arr[i], &arr[j]);  
    }  
}  
  
// Driver Code 
int main()  
{  
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};  
    int n = sizeof(arr) / sizeof(arr[0]);  
    randomize (arr, n);  
    printArray(arr, n);  
  
    return 0;  
}  
  
// This code is contributed by  
// rathbhupendra 

C

// C Program to shuffle a given array 
  
#include <stdio.h> 
#include <stdlib.h> 
#include <time.h> 
  
// A utility function to swap to integers 
void swap (int *a, int *b) 
{ 
    int temp = *a; 
    *a = *b; 
    *b = temp; 
} 
  
// A utility function to print an array 
void printArray (int arr[], int n) 
{ 
    for (int i = 0; i < n; i++) 
        printf("%d ", arr[i]); 
    printf("\n"); 
} 
  
// A function to generate a random permutation of arr[] 
void randomize ( int arr[], int n ) 
{ 
    // Use a different seed value so that we don't get same 
    // result each time we run this program 
    srand ( time(NULL) ); 
  
    // Start from the last element and swap one by one. We don't 
    // need to run for the first element that's why i > 0 
    for (int i = n-1; i > 0; i--) 
    { 
        // Pick a random index from 0 to i 
        int j = rand() % (i+1); 
  
        // Swap arr[i] with the element at random index 
        swap(&arr[i], &arr[j]); 
    } 
} 
  
// Driver program to test above function. 
int main() 
{ 
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; 
    int n = sizeof(arr)/ sizeof(arr[0]); 
    randomize (arr, n); 
    printArray(arr, n); 
  
    return 0; 
} 

Java

// Java Program to shuffle a given array 
import java.util.Random; 
import java.util.Arrays; 
public class ShuffleRand  
{ 
    // A Function to generate a random permutation of arr[] 
    static void randomize( int arr[], int n) 
    { 
        // Creating a object for Random class 
        Random r = new Random(); 
          
        // Start from the last element and swap one by one. We don't 
        // need to run for the first element that's why i > 0 
        for (int i = n-1; i > 0; i--) { 
              
            // Pick a random index from 0 to i 
            int j = r.nextInt(i+1); 
              
            // Swap arr[i] with the element at random index 
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        // Prints the random array 
        System.out.println(Arrays.toString(arr)); 
    } 
      
    // Driver Program to test above function 
    public static void main(String[] args)  
    { 
          
         int[] arr = {1, 2, 3, 4, 5, 6, 7, 8}; 
         int n = arr.length; 
         randomize (arr, n); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python

# Python Program to shuffle a given array 
import random 
  
# A function to generate a random permutation of arr[] 
def randomize (arr, n): 
    # Start from the last element and swap one by one. We don't 
    # need to run for the first element that's why i > 0 
    for i in range(n-1,0,-1): 
        # Pick a random index from 0 to i 
        j = random.randint(0,i+1) 
  
        # Swap arr[i] with the element at random index 
        arr[i],arr[j] = arr[j],arr[i] 
    return arr 
  
# Driver program to test above function. 
arr = [1, 2, 3, 4, 5, 6, 7, 8] 
n = len(arr) 
print(randomize(arr, n)) 
  
  
# This code is contributed by Pratik Chhajer 

C#

// C# Code for Number of digits  
// in the product of two numbers 
using System; 
  
class GFG 
{  
// A Function to generate a 
// random permutation of arr[] 
    static void randomize(int []arr, int n) 
    { 
        // Creating a object 
        // for Random class 
        Random r = new Random(); 
          
        // Start from the last element and 
        // swap one by one. We don't need to 
        // run for the first element  
        // that's why i > 0 
        for (int i = n - 1; i > 0; i--)  
        { 
              
            // Pick a random index 
            // from 0 to i 
            int j = r.Next(0, i+1); 
              
            // Swap arr[i] with the 
            // element at random index 
            int temp = arr[i]; 
            arr[i] = arr[j]; 
            arr[j] = temp; 
        } 
        // Prints the random array 
        for (int i = 0; i < n; i++) 
        Console.Write(arr[i] + " "); 
    } 
      
      
// Driver Code 
static void Main() 
{ 
    int[] arr = {1, 2, 3, 4,  
                 5, 6, 7, 8}; 
    int n = arr.Length; 
    randomize (arr, n); 
} 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP Program to shuffle  
// a given array 
  
// A function to generate 
// a random permutation of arr[] 
function randomize ($arr, $n) 
{ 
    // Start from the last element  
    // and swap one by one. We  
    // don't need to run for the 
    // first element that's why i > 0 
    for($i = $n - 1; $i >= 0; $i--) 
    { 
        // Pick a random index 
        // from 0 to i 
        $j = rand(0, $i+1); 
  
        // Swap arr[i] with the  
        // element at random index 
        $tmp = $arr[$i]; 
        $arr[$i] = $arr[$j]; 
        $arr[$j] = $tmp; 
    } 
    for($i = 0; $i < $n; $i++) 
    echo $arr[$i]." "; 
} 
  
// Driver Code 
$arr = array(1, 2, 3, 4,  
             5, 6, 7, 8); 
$n = count($arr); 
randomize($arr, $n); 
  
// This code is contributed by mits 
?> 

Output :

7 8 4 6 3 1 2 5

The above function assumes that rand() generates a random number.

Time Complexity: O(n), assuming that the function rand() takes O(1) time.

How does this work?
The probability that ith element (including the last one) goes to last position is 1/n, because we randomly pick an element in first iteration.

The probability that ith element goes to second last position can be proved to be 1/n by dividing it in two cases.
Case 1: i = n-1 (index of last element):
The probability of last element going to second last position is = (probability that last element doesn’t stay at its original position) x (probability that the index picked in previous step is picked again so that the last element is swapped)
So the probability = ((n-1)/n) x (1/(n-1)) = 1/n
Case 2: 0 < i < n-1 (index of non-last):
The probability of ith element going to second position = (probability that ith element is not picked in previous iteration) x (probability that ith element is picked in this iteration)
So the probability = ((n-1)/n) x (1/(n-1)) = 1/n

We can easily generalize above proof for any other position.

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

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