1419. Minimum Number of Frogs Croaking_1419. 数青蛙-优快云博客

本文链接：https://blog.youkuaiyun.com/zjucor/article/details/105752348

本文介绍了一种算法，用于确定完成给定字符串中所有'croak'叫声所需的最小青蛙数量。通过跟踪每个字符的计数并确保它们符合'croak'的正确顺序，算法能够判断字符串是否有效并计算所需青蛙的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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'.

思路：模拟一遍就好了，幸亏croak没有重复的字符。。

class Solution(object):
    def minNumberOfFrogs(self, croakOfFrogs):
        """
        :type croakOfFrogs: str
        :rtype: int
        """
        d = {}
        for c in 'croak':
            d[c] = 0
        res = 1
        for c in croakOfFrogs:
            if c == 'c':
                d['c'] += 1
                if max(d.values()) > res:
                    res += 1
            elif c == 'r':
                d['r'] += 1
                if d['r'] > d['c']:
                    return -1
            elif c == 'o':
                d['o'] += 1
                if d['o'] > d['r']:
                    return -1
            elif c == 'a':
                d['a'] += 1
                if d['a'] > d['o']:
                    return -1
            elif c == 'k':
                d['k'] += 1
                if d['k'] > d['a']:
                    return -1
                for cc in 'croak':
                    d[cc] -= 1
        return res if max(d.values()) == 0 else -1