You are 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 croaks in the given string.

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

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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: -1Explanation: The given string is an invalid combination of "croak" from different frogs.

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Constraints:

• 1 <= croakOfFrogs.length <= 105

• croakOfFrogs?is either?'c',?'r',?'o',?'a', or?'k'.

都有點(diǎn)快抑郁了，，，

1：判斷是否是按照順序，用hashmap去每次循環(huán)的時(shí)候判斷一下，（這個(gè)地方一直沒(méi)想到）

2：當(dāng)有'c'的時(shí)候 frog++,當(dāng)出現(xiàn)‘k’的時(shí)候 frog--；

3：所以就是要把這2個(gè)問(wèn)題分開(kāi)去處理，一開(kāi)始光想著一下子去處理了。所以走了不少的彎路；

4：最后就是res要max去判斷每個(gè)。

下面是代碼：

都是經(jīng)驗(yàn)教訓(xùn)啊。。。

Runtime93 ms

Beats

6.17%

Memory43.4 MB

Beats

22.14%