【leetcode】1419. Minimum Number of Frogs Croaking

题目如下：

Given the string croakOfFrogs, which represents a combination of the string "croak" from different frogs, that is, multiple frogs can croak at the same time, so multiple “croak” are mixed. Return the minimum number of different frogs to finish all the croak in the given string.

A valid "croak" means a frog is printing 5 letters ‘c’, ’r’, ’o’, ’a’, ’k’ sequentially. The frogs have to print all five letters to finish a croak. If the given string is not a combination of valid "croak" return -1. 

Example 1:

Input: croakOfFrogs = "croakcroak"
Output: 1 
Explanation: One frog yelling "croak" twice.

Example 2:

Input: croakOfFrogs = "crcoakroak"
Output: 2 
Explanation: The minimum number of frogs is two. 
The first frog could yell "crcoakroak".
The second frog could yell later "crcoakroak".

Example 3:

Input: croakOfFrogs = "croakcrook"
Output: -1
Explanation: The given string is an invalid combination of "croak" from different frogs.

Example 4:

Input: croakOfFrogs = "croakcroa"
Output: -1

Constraints:

• 1 <= croakOfFrogs.length <= 10^5
• All characters in the string are: ‘c‘, ‘r‘, ‘o‘, ‘a‘ or ‘k‘.

解题思路：从左往右遍历croakOfFrogs，遇到字符c则计数加一，出现一次完整的croak字符串则计数减一，答案就是计数出现过的最大值。

代码如下：

class Solution(object):
    def minNumberOfFrogs(self, croakOfFrogs):
        """
        :type croakOfFrogs: str
        :rtype: int
        """
        pending = {}
        lettes = [‘k‘,‘a‘,‘o‘,‘r‘,‘c‘]
        res = 0
        for char in croakOfFrogs:
            if char == ‘c‘:
                pending[char] = pending.setdefault(char,0) + 1
            else:
                pre = lettes[lettes.index(char) + 1]
                flag = False

                for key in pending.iterkeys():
                    if pre == key[-1]:
                        flag = True
                        pending[key] -= 1
                        if pending[key] == 0:
                            del pending[key]
                        if key + char != ‘croak‘:
                            pending[key+char] = pending.setdefault(key+char,0) + 1
                        break
                if flag == False:return -1

            count = 0
            for key in pending.iterkeys():
                count += pending[key]
            res = max(res,count)

        # uncomplete string
        if len(pending) > 0:
            return -1
        return res

【leetcode】1419. Minimum Number of Frogs Croaking

原文：https://www.cnblogs.com/seyjs/p/13026424.html