Finding all sequences of A, B such that have a specified number of each element

For example, given the two letters A and B, I'd like to generate all strings of length n that have x A's and y B's.

I'd like this to be done efficiently. One way that I've considered is to build a length x list of A's, and then insert y B's into the list every possible way. But insertion into a python list is linear, so this method would suck as the list gets big.

PERFORMANCE GOAL (this may be unreasonable, but it is my hope): Generate all strings of length 20 with equal numbers of A and B in time less than a minute.

EDIT: Using permutations('A' * x, 'B' * y) has been suggested. While not a bad idea, it's wasting a lot. If x = y = 4, you'd generate the string 'AAAABBBB' many times. Is there a better way that might generate each string only once? I've tried code to the effect of set(permutations('A' * x, 'B' * y)) and it is too slow.

Solution

Regarding your concerns with the performance, here is an actual generator implementation of your idea (without insert). It finds the positions for B and fill the list accordingly.

import itertools

def make_sequences(num_a, num_b):
    b_locations = range(num_a+1)
    for b_comb in itertools.combinations_with_replacement(b_locations, num_b):
        result = []
        result_a = 0
        for b_position in b_comb:
            while b_position > result_a:
                result.append('A')
                result_a += 1
            result.append('B')
        while result_a < num_a:
            result.append('A')
            result_a += 1
        yield ''.join(result)

It does perform better. Comparing with the Greg Hewgill's solution (naming it make_sequences2):

In : %timeit list(make_sequences(4,4))
10000 loops, best of 3: 145 us per loop

In : %timeit make_sequences2(4,4)
100 loops, best of 3: 6.08 ms per loop

Edit

A generalized version:

import itertools

def insert_letters(sequence, rest):
    if not rest:
        yield sequence
    else:
        letter, number = rest[0]
        rest = rest[1:]
        possible_locations = range(len(sequence)+1)
        for locations in itertools.combinations_with_replacement(possible_locations, number):
            result = []
            count = 0
            temp_sequence = sequence
            for location in locations:
                while location > count:
                    result.append(temp_sequence[0])
                    temp_sequence = temp_sequence[1:]
                    count += 1
                result.append(letter)
            if temp_sequence:
                result.append(temp_sequence)
            for item in insert_letters(''.join(result), rest):
                yield item

def generate_sequences(*args):
    '''
    arguments : squence of (letter, number) tuples
    '''
    (letter, number), rest = args[0], args[1:]
    for sequence in insert_letters(letter*number, rest):
        yield sequence

Usage:

for seq in generate_sequences(('A', 2), ('B', 1), ('C', 1)):
    print seq

# Outputs
# 
# CBAA
# BCAA
# BACA
# BAAC
# CABA
# ACBA
# ABCA
# ABAC
# CAAB
# ACAB
# AACB
# AABC