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array 05: is crypt solution(reduce,map,any,all)


1. 题目

A cryptarithm is a mathematical puzzle for which the goal is to find the correspondence between letters and digits, such that the given arithmetic equation consisting of letters holds true when the letters are converted to digits.

You have an array of strings crypt, the cryptarithm, and an an array containing the mapping of letters and digits, solution. The array crypt will contain three non-empty strings that follow the structure: [word1, word2, word3], which should be interpreted as the word1 + word2 = word3 cryptarithm.

If crypt, when it is decoded by replacing all of the letters in the cryptarithm with digits using the mapping in solution, becomes a valid arithmetic equation containing no numbers with leading zeroes, the answer is true. If it does not become a valid arithmetic solution, the answer is false.

Example

  • For crypt = ["SEND", "MORE", "MONEY"] and

    solution = [['O', '0'],
              ['M', '1'],
              ['Y', '2'],
              ['E', '5'],
              ['N', '6'],
              ['D', '7'],
              ['R', '8'],
              ['S', '9']]

    the output should be
    isCryptSolution(crypt, solution) = true;
    When you decrypt "SEND", "MORE", and "MONEY" using the mapping given in crypt, you get 9567 + 1085 = 10652 which is correct and a valid arithmetic equation.

  • For crypt = ["TEN", "TWO", "ONE"] and

    solution = [['O', '1'],
              ['T', '0'],
              ['W', '9'],
              ['E', '5'],
              ['N', '4']]

    the output should be
    isCryptSolution(crypt, solution) = false;
    Even though 054 + 091 = 145, 054 and 091 both contain leading zeroes, meaning that this is not a valid solution.

Input/Output

  • [time limit] 4000ms (py)

  • [input] array.string crypt
    An array of three non-empty strings containing only uppercase English letters.
    Guaranteed constraints: crypt.length = 3, 1 ≤ crypt[i].length ≤ 14.

  • [input] array.array.char solution
    An array consisting of pairs of characters that represent the correspondence between letters and numbers in the cryptarithm. The first character in the pair is an uppercase English letter, and the second one is a digit in the range from 0 to 9.
    Guaranteed constraints:
    solution[i].length = 2,
    'A' ≤ solution[i][0] ≤ 'Z',
    '0' ≤ solution[i][1] ≤ '9',
    solution[i][0] ≠ solution[j][0], i ≠ j,
    solution[i][1] ≠ solution[j][1], i ≠ j.
    It is guaranteed that solution only contains entries for the letters present in crypt and that different letters have different values.

  • [output] boolean
    Return true if the solution represents the correct solution to the cryptarithm crypt, otherwise return false.


2. 解法

个人解法

def isCryptSolution(crypt, solution):
    def wordTransfer(word, sl_dict):
        nums = ""
        for char in word:
            nums += sl_dict[char]
        return "" if len(nums) > 1 and nums[0] == "0" else int(nums)

    sl_dict = {x[0]: x[1] for x in solution}
    crypt_trans = [wordTransfer(x, sl_dict) for x in crypt]

    if "" in crypt_trans:
        return False
    else:
        if crypt_trans[0] + crypt_trans[1] == crypt_trans[2]:
            return True
        else:
            return False

python2精品解法

def isCryptSolution(crypt, solution):
    sl_dict = {char: num for char, num in solution}
    crypt_trans = [reduce(lambda x,y: x+y,map(lambda x: sl_dict[x], word)) for word in crypt]

    return not any(x != "0" and x[0] == "0" for x in crypt_trans) and \
        int(crypt_trans[0]) + int(crypt_trans[1]) == int(crypt_trans[2])

python2版本没有精品解法,是根据下面的python3精品解法改编来。
重点1:在于嵌套的map和reduce函数
重点2:在于any和all函数

python3精品解法

def isCryptSolution(crypt, solution):
    dic = {ord(c): d for c, d in solution}
    *v, = map(lambda x: x.translate(dic), crypt)
    return not any(x != "0" and x.startswith("0") for x in v) and \
        int(v[0]) + int(v[1]) == int(v[2])

python3 版本,python2不能用,是因为codefights 不支持import操作,如果要完全模仿py3的写法,需要import string