You can pair socks efficiently using the Merge Sort algorithm, which sorts the socks by size, type, or color. This ensures that similar socks are placed next to each other, allowing for quick pairing.

Key takeaways:

The problem of pairing socks from a random pile can be efficiently solved using the merge sort algorithm, which organizes the socks by their size, type, or color, making it easier to pair alike socks.

Merge sort is a divide-and-conquer algorithm that recursively divides the array (or pile of socks) into smaller subarrays, sorts them, and then merges them back into a sorted array. The process ensures all socks are placed next to their pair.

The time complexity of the merge sort algorithm is

`O(N*log(N)),`

which makes it an optimal solution for large piles of socks or arrays.Merge sort requires

`O(N)`

auxiliary space to store the divided arrays, making it a bit costly in terms of memory usage, but still manageable for moderate-sized arrays.

Given a pile of socks with random distribution, we’ll find the most efficient way of pairing up the alike socks.

We model this problem by assuming an array `Pile`

with `N`

unique elements with exactly two instances of each element. We arrange these elements in a `Pile`

so that each `N`

element lies next to its partner sock. This leads to a total of `2N`

elements in the `Pile`

.

The working of merge sort to pair socks from a given pile is as shown in the below slides:

We use the **merge sort **algorithm to produce the optimized solution for solving this problem with time complexity as follows:

Merge sort is a recursive algorithm that breaks the given array into subarrays and sorts them. Then it keeps combining these sorted subarrays to produce larger sorted arrays. This process is repeated until an array of the original length is recovered.

In our socks pairing problem, once we sort the socks based on their sizes, types, or colors, those that are alike will be placed next to each other. In this manner, all the socks will be paired up.

Let's look at the code below for this solution:

def Merge(Pile):if len(Pile)<2:returnelse:m = len(Pile)//2 #array's middle elementRight = Pile[m:] #array's right halfLeft = Pile[:m] #array's left halfMerge(Right)Merge(Left)i = j = k = 0while i < len(Left) and j < len(Right):if Left[i] <= Right[j]:Pile[k] = Left[i]i += 1else:Pile[k] = Right[j]j += 1k += 1while i < len(Left):Pile[k] = Left[i]i += 1k += 1while j < len(Right):Pile[k] = Right[j]j += 1k += 1if __name__ == '__main__':Pile = [4, 2, 7, 2, 4, 1, 1, 7]print("Unpaired socks: ", Pile)Merge(Pile)print("Paired socks: ", Pile)

**Lines 2–3:**We define the base case for recursion. The function will return if the array has just one or no elements.**Lines 5–7:**Here we locate the middle element of the array and split it into left and right halves.**Lines 14–21:**We a loop with two iterators that move over`Left`

and`Right`

subarrays, comparing their elements and merging the smaller ones into`Arr`

.**Lines 23–31:**We check if there is still any element left in the`Left`

or`Right`

subarrays to merge into`Arr`

.

The time complexity of this recursive algorithm is *O(N*log(N)).* It uses auxiliary space for storing the array elements then its space complexity is *O(N)*.

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