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.

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 <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); ` ` ` ` ` `// 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) ` ` ` ` ` `# 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); ` ` ` ` ` `// 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` `); ` ` ` ` ` `// 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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