如何在不排序列表的情况下找到随机列表中两个数字之间的最短距离。 我的代码如下，我需要另一种快速的方式

Question

    
import random



L = [random.randrange(1, 100**7) for i in range(100*100)]
       
#it creates a random list

for x in L:
    min = L[0]
    for y in L:
        if x != y and abs(x-y) < min:
            min = abs(x-y)
**#it basically cheeks every element with every one and update the min value id new one is found**


print(min)

any faster way doing so without sorting? someone suggested it can be done using a hashing table but I cant relate to it so pls help me with the logic and connection of this question with hashing table

Answer 1

The numbers can be stored in nested lists according with its binary (or whatever) representation. We use a fix digits number (for this example 3). For example 2 will be represented as 010

Then we put an element in

numbers[0][1][0]

The tree with just the number 2 would be:

[
    [None, [True, None]], None
]

If we add number 3 (011):

[
    [None, [True, True]], None
]

add number 1 (001):

[
    [[None, True], [True, True]], None
]

All the lists has 2 elements and are created as needed. Unused elelements are set to 'None'

We actually dont need to put the number because all the information is in the indices. So we put just True to say that the number 2 is in the tree.

Now to find the closer number in the tree to some other number just do a binary search.

Now compare the minimal distance with the stored minimal distance , update the stored minimal distance if needed and store the new number in the tree.

I think this O(N * logN). The log comes from the binary storage and search.