Shuffle a given array

Software Development
#1

Hello Everyone,

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++ 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;

}

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.