Prefix Flip (Easy Version) (Codeforces 1381A1)
Tags: Others, strings

This is the easy version of the problem. The difference between the versions is the constraint on $$$n$$$ and the required number of operations. You can make hacks only if all versions of the problem are solved.There are two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$ (a binary string is a string consisting of symbols $$$0$$$ and $$$1$$$). In an operation, you select a prefix of $$$a$$$, and simultaneously invert the bits in the prefix ($$$0$$$ changes to $$$1$$$ and $$$1$$$ changes to $$$0$$$) and reverse the order of the bits in the prefix.For example, if $$$a=001011$$$ and you select the prefix of length $$$3$$$, it becomes $$$011011$$$. Then if you select the entire string, it becomes $$$001001$$$.Your task is to transform the string $$$a$$$ into $$$b$$$ in at most $$$3n$$$ operations. It can be proved that it is always possible.

Input: The first line contains a single integer $$$t$$$ ($$$1\le t\le 1000$$$)  — the number of test cases. Next $$$3t$$$ lines contain descriptions of test cases.The first line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 1000$$$)  — the length of the binary strings.The next two lines contain two binary strings $$$a$$$ and $$$b$$$ of length $$$n$$$.It is guaranteed that the sum of $$$n$$$ across all test cases does not exceed $$$1000$$$.

Output: For each test case, output an integer $$$k$$$ ($$$0\le k\le 3n$$$), followed by $$$k$$$ integers $$$p_1,\ldots,p_k$$$ ($$$1\le p_i\le n$$$). Here $$$k$$$ is the number of operations you use and $$$p_i$$$ is the length of the prefix you flip in the $$$i$$$-th operation.

Note: In the first test case, we have $$$01\to 11\to 00\to 10$$$.In the second test case, we have $$$01011\to 00101\to 11101\to 01000\to 10100\to 00100\to 11100$$$.In the third test case, the strings are already the same. Another solution is to flip the prefix of length $$$2$$$, which will leave $$$a$$$ unchanged.

Sample Inputs:
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1

Sample Outputs:
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Solution (Python):
import sys
input = sys.stdin.readline
T = int(input())
for _ in range(T):
    n = int(input())
    a = input()
    b = input()
    scheck = []
    pos = 0
    pos1 = n - 1
    reverse = 0
    for j in range(n - 1, -1, -1):
        if not reverse:
            if a[pos] != b[j]:
                scheck.append(j + 1)
            else:
                scheck.append(1)
                scheck.append(j + 1)
            pos += 1
            reverse = 1
        else:
            if a[pos1] != b[j]:
                scheck.append(1)
                scheck.append(j + 1)
            else:
                scheck.append(j + 1)
            pos1 -= 1
            reverse = 0
    print(len(scheck), *scheck)

Solution (Python):
for _ in range(int(input())):
    n = int(input())
    a = list(input())
    b = list(input())
    result = []
    for i in range(n, 0, -1):
        if a[i - 1] != b[i - 1]:
            result += [i, 1, i]
    print(len(result), *result)

Solution (Python):
def solve():
    n = int(input())
    a = input().strip()
    b = input().strip()
    ans = []
    now = int(a[0])
    pos = 1
    while pos < n:
        if int(a[pos]) == now:
            pos += 1
        else:
            ans.append(pos)
            pos += 1
            now = 1 - now
    pos = n
    while pos >= 1:
        if int(b[pos - 1]) == now:
            pos -= 1
        else:
            ans.append(pos)
            pos -= 1
            now = 1 - now
    print(len(ans), end=' ')
    if len(ans) == 0:
        print()
    else:
        for i in range(len(ans)):
            print(ans[i], end=' ')
        print()

def main():
    t = int(input())
    for _ in range(t):
        solve()
if __name__ == '__main__':
    main()