MicroSoft interview 2022 test1

def solution(S, B):
    # write your code in Python (Python 3.6)
    fixed = 0
    holes = []
    len_hole = 0
    for idx in range(len(S)):
        # print(S[idx])
        if S[idx] == 'x':
            len_hole += 1
        else:
            if len_hole > 0:
                holes.append(len_hole)
                len_hole = 0
    if len_hole > 0:
        holes.append(len_hole)
    # print("1:", holes)
    holes = sorted(holes)
    # print("2:", holes)
    for item in holes[::-1]:
        cost = item + 1
        if B >= cost:
            B -= cost
            fixed += item
        elif 1 < B < cost:
            fixed += B - 1
            B = 0
        else:
            break
    return fixed


if __name__ == '__main__':
    # A = 24
    S = '...xxx..x....xxx.'
    B = 7
    lea_count = solution(S, B)
    print(lea_count)

def solution(S):
    # write your code in Python (Python 3.6)
    min_swap = 0
    r_swap, l_swap = [], []
    sub_s = []
    start, end = 0, 0
    for idx, item in enumerate(S):
        if item == 'R':
            start = idx
            break
    for idx, item in enumerate(S[::-1]):
        if item == 'R':
            end = len(S) - idx
            break
    # print(start, end)
    S = S[start:end]
    # print(S)
    r_count = 0
    w_count = 0
    for item in S:
        if item == 'R':
            r_count += 1
        else:
            w_count += 1
            l_swap.append(r_count)
    r_count = 0
    for item in S[::-1]:
        if item == 'R':
            r_count += 1
        else:
            r_swap.append(r_count)
    r_swap = r_swap[::-1]
    for l, r in zip(l_swap, r_swap):
        min_swap += min(l, r)
        if min_swap > 1e9:
            min_swap = -1
            break

    return min_swap


if __name__ == '__main__':
    # A = 24
    S = 'WRW' * 10000
    B = 7
    lea_count = solution(S)
    print(lea_count)

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

# def solution(A, B, S):
#     # write your code in Python (Python 3.6)
#     if len(set(A + B)) < len(A):
#         return False
#     fixed = [0] * S
#     flag = True
#     while flag:
#         for i in range(len(A) - 1):
#             for j in range(i + 1, len(A)):
#                 if A[i] == A[j]:
#                     if fixed[i] == 0:
#                         fixed[i] = 1
#                         A[i] = B[i]
#                     else:
#                         return False
#         flag = False
#     # print(A)
#     return True

def solution(A, B, S):
    # write your code in Python (Python 3.6)
    slices = set(A + B)
    if len(slices) >= len(A):
        assigned = True
    else:
        assigned = False
    # print(A)
    # a = ["a", "b", "a", "c", "a", "c", "b", "d", "e", "c", "a", "c"]
    # # set集合去重
    # duixiang = set(a)  # 先去重，取出计数对象
    # # 保存为dict，一一对应
    # d = {}
    # for i in duixiang:
    #     d[i] = a.count(i)
    # # 对字典按value排序
    # a = sorted(d.items(), key=lambda x: x[1], reverse=True)
    return assigned


if __name__ == '__main__':
    # False
    A = [1, 2, 4, 2, 4, 6, 7, 5]
    B = [2, 1, 3, 7, 5, 5, 6, 4]
    S = 8

    # True
    # A = [1, 2, 2]
    # B = [3, 1, 3]
    # S = 3

    # True
    # A = [1, 2, 2, 8, 4, 6, 7, 6]
    # B = [2, 1, 3, 7, 5, 5, 6, 4]
    # S = 8

    # True
    # A = [1, 1, 4, 8, 4, 6, 7, 6]
    # B = [2, 2, 3, 7, 5, 5, 6, 4]
    # S = 8

    # True
    # A = [1] * 100000
    # B = [(i + 1) % 100001 for i in range(1, 100000)] + [1000]
    # S = 100000

    lea_count = solution(A, B, S)
    print(lea_count)