Initial Thoughts
Normally I would display more than two approaches, but shuffling is deceptively easy to do almost properly, and the FisherYates algorithm is both the canonical solution and asymptotically optimal.
A few notes on randomness are necessary before beginning  both approaches displayed below assume that the languages' pseudorandom number generators (PRNGs) are sufficiently random. The sample code uses the simplest techniques available for getting pseudorandom numbers, but for each possible permutation of the array to be truly equally likely, more care must be taken. For example, an array of length has distinct permutations. Therefore, in order to encode all permutations in an integer space, bits are necessary, which may not be guaranteed by the default PRNG.
Approach #1 Brute Force [Accepted]
Intuition
If we put each number in a "hat" and draw them out at random, the order in which we draw them will define a random ordering.
Algorithm
The brute force algorithm essentially puts each number in the aforementioned
"hat", and draws them at random (without replacement) until there are none
left. Mechanically, this is performed by copying the contents of array
into
a second auxiliary array named aux
before overwriting each element of
array
with a randomly selected one from aux
. After selecting each random
element, it is removed from aux
to prevent duplicate draws. The
implementation of reset
is simple, as we just store the original state of
nums
on construction.
The correctness of the algorithm follows from the fact that an element
(without loss of generality) is equally likely to be selected during all
iterations of the for
loop. To prove this, observe that the probability of a
particular element being chosen on the th iteration (indexed from 0)
is simply being chosen during the th iteration not being
chosen before the th iteration. Given that the array to be shuffled has
elements, this probability is more concretely stated as the following:
When expanded (and rearranged), it looks like this (for sufficiently large ):
For the base case (), it is trivial to see that . For , the numerator of each fraction can be cancelled with the denominator of the next, leaving the from the 0th draw as the only uncancelled denominator. Therefore, no matter on which draw an element is drawn, it is drawn with a chance, so each array permutation is equally likely to arise.
Complexity Analysis

Time complexity :
The quadratic time complexity arises from the calls to
list.remove
(orlist.pop
), which run in linear time. linear list removals occur, which results in a fairly easy quadratic analysis. 
Space complexity :
Because the problem also asks us to implement
reset
, we must use linear additional space to store the original array. Otherwise, it would be lost upon the first call toshuffle
.
Approach #2 FisherYates Algorithm [Accepted]
Intuition
We can cut down the time and space complexities of shuffle
with a bit of
cleverness  namely, by swapping elements around within the array itself, we
can avoid the linear space cost of the auxiliary array and the linear time
cost of list modification.
Algorithm
The FisherYates algorithm is remarkably similar to the brute force solution. On each iteration of the algorithm, we generate a random integer between the current index and the last index of the array. Then, we swap the elements at the current index and the chosen index  this simulates drawing (and removing) the element from the hat, as the next range from which we select a random index will not include the most recently processed one. One small, yet important detail is that it is possible to swap an element with itself  otherwise, some array permutations would be more likely than others. To see this illustrated more clearly, consider the animation below:
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Complexity Analysis

Time complexity :
The FisherYates algorithm runs in linear time, as generating a random index and swapping two values can be done in constant time.

Space complexity :
Although we managed to avoid using linear space on the auxiliary array from the brute force approach, we still need it for
reset
, so we're stuck with linear space complexity.
Analysis written by: @emptyset