Shuffle a given array using Fisher–Yates shuffle Algorithm

27 Jun, 2022
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 be equally likely. 
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 a random number in O(1) time. The idea is to start from the last element and swap it with a randomly selected element from the whole array (including the 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 that is as follows:  
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]
Flowchart:
flowchart
Following is an implementation of this algorithm.
C++
C
Java
Python3
C#
PHP
Javascript
// 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 integersvoid swap (int *a, int *b){    int temp = *a;    *a = *b;    *b = temp;} // A utility function to print an arrayvoid 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 Codeint 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 Program to shuffle a given array #include <stdio.h>#include <stdlib.h>#include <time.h> // A utility function to swap to integersvoid swap (int *a, int *b){    int temp = *a;    *a = *b;    *b = temp;} // A utility function to print an arrayvoid 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 Program to shuffle a given arrayimport 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 Program to shuffle a given arrayfrom random import randint # 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 = 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# Code for Number of digits// in the product of two numbersusing 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 Codestatic 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 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?>
<script>// JavaScript Program to shuffle a given array // A function to print an arraylet printArray = (arr, n)=>{    ans = '';    for (let i = 0; i < n; i++)    {        ans += arr[i] + " ";    }    console.log(ans);} // A function to generate a random// permutation of arrlet 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 (let i = n - 1; i > 0; i--)    {             // Pick a random index from 0 to i inclusive        let j = Math.floor(Math.random() * (i + 1));         // Swap arr[i] with the element        // at random index        [arr[i], arr[j]] = [arr[j], arr[i]];    }} // Driver Codelet arr = [1, 2, 3, 4, 5, 6, 7, 8];let n = arr.length;randomize (arr, n);printArray(arr, n); // This code is contributed by rohitsingh07052.</script>
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., Auxiliary Space: O(1)
How does this work? 
The probability that ith element (including the last one) goes to the last position is 1/n, because we randomly pick an element in the first iteration.The probability that ith element goes to the second last position can be proved to be 1/n by dividing it into 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/nWe can easily generalize above proof for any other position.
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