[プログラマー](python)秘密地図-2018 KAKAO BLIND RECRUIMENT

🧩 首都コード

🏁 説明する

def solution(n,arr1,arr2):

    new_arr1 = []
    new_arr2 = []
    sum_arr = []

    for i in range(len(arr1)):
        new_arr1.append(format(arr1[i], 'b'))
        if len(new_arr1[i]) < n:
            new_arr1[i] = (n-len(new_arr1[i]))*'0'+new_arr1[i]
        new_arr2.append(format(arr2[i], 'b'))
        if len(new_arr2[i]) < n:
            new_arr2[i] = (n-len(new_arr2[i]))*'0'+new_arr2[i]


    for i in range(len(new_arr1)):
        for j in range(len(new_arr2)):
            if new_arr1[i][j] == '0' and new_arr2[i][j] == '0':
                sum_arr.append(' ')
            elif new_arr1[i][j] or new_arr2[i][j] == '1':
                sum_arr.append('#')

    sum_arr = ''.join(sum_arr)
    answer = list(map(''.join, zip(*[iter(sum_arr)]*n)))
    
    return answer

📌 环保形式的复印

📌 ''.join()文字列的标志,文字列的唇膏

🧩 別の解釈

def solution(n, arr1, arr2):
    answer = []
    for i,j in zip(arr1,arr2):
        a12 = str(bin(i|j)[2:])
        a12=a12.rjust(n,'0')
        a12=a12.replace('1','#')
        a12=a12.replace('0',' ')
        answer.append(a12)
    return answer

📌 bin(i|j)

new_arr1.append(format(arr1[i], 'b'))

📌 rjust()

if len(new_arr1[i]) < n:
            new_arr1[i] = (n-len(new_arr1[i]))*'0'+new_arr1[i]

📌 replace()

🧩 その他の解釈。

solution = lambda n, arr1, arr2: ([''.join(map(lambda x: '#' if x=='1' else ' ', "{0:b}".format(row).zfill(n))) for row in (a|b for a, b in zip(arr1, arr2))])