I have a list of 5 letters [‘A’, ‘B’, ‘N’, ‘M’,’E’].
I want to print all the words (word means a sequence of letters, it doesn’t have to be a valid English word) of length 10 letters that have exactly two letters A. Order is important.
I have tried with itertools.product as it appeared to be the most promising solution:
from itertools import product
letters = ['A', 'B', 'N', 'M','E']
for word in product(letters, repeat=10):
res = ''.join(str(x) for x in word)
print(res)
The problem with this approach is that I can’t really control the number of occurrences of the letter A as it returns the word composed of 10 letters of A.
Is there a solution for this? Thanks
EDIT 1 Example of possible words: BANAMEMNEB : it has only twice the letter A, we don’t care about other letters.
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Answer
The way to do it efficiently is to iterate over all combinations of the positions of the letters ‘A’, and for each such combination – iterate over all the ways other letters can be positioned, and just insert the ‘A’s there.
Be warned that with your inputs this will produce almost 3 million words!
On my machine it was printing the words for so long that I had to manually stop execution.
So here is the smaller example:
letters = ['A', 'b', 'c']
num_As = 2
num_total = 4
from itertools import combinations, product
for indices_A in combinations(range(num_total), num_As):
for rest in product(letters[1:], repeat=num_total - num_As):
rest = list(rest)
for index_A in indices_A:
rest.insert(index_A, 'A')
print(''.join(rest))
AAbb AAbc AAcb AAcc AbAb AbAc AcAb AcAc AbbA AbcA AcbA AccA bAAb bAAc cAAb cAAc bAbA bAcA cAbA cAcA bbAA bcAA cbAA ccAA