299.Bulls and Cows

Tags: [map]

Link: https://leetcode.com/problems/bulls-and-cows/\#/description

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called "bulls") and how many digits match the secret number but locate in the wrong position (called "cows"). Your friend will use successive guesses and hints to eventually derive the secret number.

For example:

Secret number:  "1807"
Friend's guess: "7810"

Hint:1bull and3cows. (The bull is8, the cows are0,1and7.)

Write a function to return a hint according to the secret number and friend's guess, use Ato indicate the bulls andBto indicate the cows. In the above example, your function should return"1A3B".

Please note that both secret number and friend's guess may contain duplicate digits, for example:

Secret number:  "1123"
Friend's guess: "0111"

In this case, the 1st 1in friend's guess is a bull, the 2nd or 3rd1is a cow, and your function should return"1A1B".

You may assume that the secret number and your friend's guess only contain digits, and their lengths are always equal.

Solution: Map

class Solution(object):
    def getHint(self, secret, guess):
        """
        :type secret: str
        :type guess: str
        :rtype: str
        """
        if not secret or not guess:
            return '0A0B'

        secret_char_map = {}
        for c in secret:
            if c not in secret_char_map:
                secret_char_map[c] = 1
            else:
                secret_char_map[c] += 1

        guess_char_map = {}
        for c in guess:
            if c not in guess_char_map:
                guess_char_map[c] = 1
            else:
                guess_char_map[c] += 1

        num_of_bulls = 0
        for i in xrange(len(secret)):
            sc = secret[i]
            gc = guess[i]

            if sc == gc:
                num_of_bulls += 1
                secret_char_map[sc] -= 1
                if not secret_char_map[sc]:
                    del secret_char_map[sc]

                guess_char_map[gc] -= 1
                if not guess_char_map[gc]:
                    del guess_char_map[gc]

        num_of_cows = 0
        for gc in guess_char_map:
            if gc in secret_char_map:
                num_of_cows += min(secret_char_map[gc], guess_char_map[gc])

        return '{}A{}B'.format(num_of_bulls, num_of_cows)

Revelation:

• In the last for loop, num_of_cows += min(secret_char_map[gc], guess_char_map[gc]), because we need find the 'intersection'.

Note:

• Time complexity = O(n), n is the length of given secret or guess (length of secret is equal to the length of guess).