Reducing time complexity in sorting two lists - list

Question

I have two lists: list_1 and list_2 containing 32 bit - strings and each is of length 2^12.
I want to sort these lists to get those pairs which contain equal bits at 20 specified positions.
list_w = []
for i in range(0, 4096):
for j in range(0, 4096):
if list_1[i][12:32] == list_2[j][12:32]:
value = (list_1[i], list_2[j])
list_w.append(value)
else:
continue
I have assumed 20 positions at the end and sort all 2^24 pairs. But this takes too much time. The answer was we can do this in 2^12 time by sorting or hashing how?

Answer 1

Use a dictionary to map the values at these 20 positions to the list of strings that match those values at those positions.
Building a dict of lists is a very classical idiom in python:
d1 = {}
for s in list_1:
d1.setdefault(s[12:32], []).append(s)
d2 = {}
for s in list_2:
d2.setdefault(s[12:32], []).append(s)
Then you can find your pairs by iterating on the keys that are in both dictionaries; i.e., keys that are in the intersection of the two sets of keys.
list_w = [
(s1, s2)
for k in set(d1.keys()).intersection(d2.keys())
for s1 in d1[k] for s2 in d2[k]
]
Testing:
list_1 = ['abcdefghijklmnopqrstuvwxyzabcdef', 'aaaaaaaaaaaamnopqrstuvwxyzabcdef', 'aaaaaaaaaaaabbbbbbbbbbbbbbbbbbbb']
list_2 = ['ABCDEFGHIJKLmnopqrstuvwxyzabcdef', 'bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb']
...
print(d1)
# {'mnopqrstuvwxyzabcdef': ['abcdefghijklmnopqrstuvwxyzabcdef',
# 'aaaaaaaaaaaamnopqrstuvwxyzabcdef'],
# 'bbbbbbbbbbbbbbbbbbbb': ['aaaaaaaaaaaabbbbbbbbbbbbbbbbbbbb']}
print(d2)
# {'mnopqrstuvwxyzabcdef': ['ABCDEFGHIJKLmnopqrstuvwxyzabcdef'],
# 'bbbbbbbbbbbbbbbbbbbb': ['bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb']}
print(list_w)
# [('aaaaaaaaaaaabbbbbbbbbbbbbbbbbbbb', 'bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb'),
# ('abcdefghijklmnopqrstuvwxyzabcdef', 'ABCDEFGHIJKLmnopqrstuvwxyzabcdef'),
# ('aaaaaaaaaaaamnopqrstuvwxyzabcdef', 'ABCDEFGHIJKLmnopqrstuvwxyzabcdef')]