1419. Minimum Number of Frogs Croaking

Problem Description

Your task is to find the minimum number of different frogs that could have produced a given string croakOfFrogs. Each frog croaks in a specific sequence 'c', 'r', 'o', 'a', 'k'. It's possible for multiple frogs to croak at the same time, which results in an interleaving of these sequences. The goal is to find out how many distinct frogs are needed to generate the string as it's given, with the stipulation that a valid "croak" sound from a frog is represented by the characters "croak" appearing sequentially. If the string does not represent a combination of valid "croak" sounds, the function should return -1.

Intuition

To solve this problem, we need to track the progress of each croak as it's being formed.

Firstly, since each croak is composed of 5 characters, if the total length of croakOfFrogs isn't a multiple of 5, it's impossible to form croaks with equal numbers of 'c's, 'r's, 'o's, 'a's and 'k's and we can immediately return -1.

Next, we create a counter array cnt that keeps track of how many frogs are currently at each stage of croaking. The index in the array corresponds to the characters in the order 'c', 'r', 'o', 'a', 'k', initialized as [0, 0, 0, 0, 0].

Now, we iterate through the characters in croakOfFrogs and update the cnt array for the stages:

1. Every time we encounter a 'c', we start a new croak, hence we increment the count for 'c'.

2. For the other characters ('r', 'o', 'a', 'k'), we check if there's a preceding character sound that's in progress (the previous character count should be greater than 0). If not, the string is invalid and we return -1.

3. If a preceding character is found, we decrement the counter for that character (since the frog has moved to the next stage of croaking) and increment the counter for the current character.

To find the minimum number of frogs needed, we keep a counter x which increments every time we start a "croak" with 'c' and decrement when a croak is completed with 'k'.

The answer will be the maximum value of x at any point, which represents the peak number of concurrent croaking frogs required to generate the given string.

Finally, we check if x is 0 after the iteration, which would mean all croaks are completed. If any croaks are left unfinished (x is not 0), we return -1 as the string wouldn't be valid in that case. Otherwise, we return the maximum number recorded in x as the answer.

The key insight here is ensuring that the sequence of 'croak' is valid and tracking the peak concurrency, which equates to the minimum number of different frogs required.

Solution Approach

The implementation follows a straightforward approach using a list to keep track of the stages of croaking for multiple frogs simultaneously.

1. Initialization: A dictionary idx is created to map each character 'c', 'r', 'o', 'a', 'k' to an index 0 through 4. This maps each character to its respective position in the croaking sequence. Also, an array cnt of 5 zeroes is initialized to count the number of frogs at each stage of croaking.

2. Check Length: The first if-statement checks if the length of croakOfFrogs is a multiple of 5 (since each complete croak sequence has 5 letters). If not, the function returns -1 immediately because it's impossible to form valid croak sequences otherwise.

3. Iteration: The algorithm iterates over each character in croakOfFrogs. Using map(idx.get, croakOfFrogs), each character is replaced by its corresponding index. This allows the function to work with indices instead of characters, facilitating easier increment/decrement operations.

4. Counting Progress: For each character index i:

   • If we encounter a 'c' (i == 0), it represents the start of a new croak, so the count of 'c' (cnt[0]) is incremented, and the variable x is also incremented. The variable ans keeps track of the maximum value that x takes, representing the peak concurrency of croaking frogs.
   • For indices 1 to 4 (r, o, a, k), the algorithm checks if there's a preceding character count that's not zero (which would indicate that a frog is in the previous stage of croaking). If such a count does not exist (cnt[i - 1] == 0), it means the sequence is incomplete or out of order, so -1 is returned.
   • If a valid sequence is maintained, the counter for the preceding stage is decremented (cnt[i - 1] -= 1), indicating that one frog has moved on to the current stage.
   • When a 'k' is encountered (i == 4), it signals the end of a croak, so x is decremented.

5. Final Check: After the loop, if x is not 0, it means there are incomplete croak sequences, and thus the input string is invalid. The function returns -1 in this case. If x is 0, the function returns ans, which contains the maximum number of frogs that were croaking at the same time at any point during the input string, which is the answer to the problem.

The core algorithm hinges on the idea of maintaining the sequential integrity of the "croak" sounds and tracking the maximum number of overlapping croaks. This is done by incrementing and decrementing counters in a list based on the sequential sound stages of the frogs.

This approach effectively uses constant space (since the size of cnt is always five) and linear time complexity relative to the length of the input string.

Example Walkthrough

Let's illustrate the solution approach using a small example.

Consider the input string croakOfFrogs as "croakcroa".

1. Initialization: We start by mapping each character ('c', 'r', 'o', 'a', 'k') to its index (0 - 4) and initialize the cnt counter array as [0, 0, 0, 0, 0].

2. Check Length: The length of "croakcroa" is 9, which is not a multiple of 5. Therefore, by our algorithm, this string cannot be made up of a valid sequence of "croak" sounds. As per the approach, we should return -1 immediately.

However, for illustrative purposes, let's pretend that the string was "croakcroak" to show how the process would continue with a valid string length.

3. Iteration: Now we map the characters to their respective indices as we iterate through "croakcroak", giving us the indices [0, 1, 2, 3, 4, 0, 1, 2, 3, 4].

4. Counting Progress: We process the indexed characters in order:

   • For the first 'c' (i == 0), we increment cnt[0] to 1 and also increment x to 1. The ans is now 1.
   • Next, we read 'r' (i == 1), we see that cnt[0] is 1 (since there is an ongoing 'c' croak), so we decrement cnt[0] to 0 and increment cnt[1] to 1. No change in x or ans.
   • This process continues for 'o', 'a', and 'k', with the following states: cnt becomes [0, 0, 0, 0, 0], and x comes back to 0 each time after reading 'k'.
   • Upon the next 'c', we again increment cnt[0] to 1 and x to 1, pushing ans to 2 because we have two concurrent croaks.
   • As we continue, cnt eventually returns to [0, 0, 0, 0, 0] and x to 0, after finishing the second "croak".

5. Final Check: We finished processing the string with x being 0, confirming all "croak" sounds are complete. ans is 2, meaning at one point during the process, there were two frogs croaking at the same time. This is the minimum number of frogs required to produce the sequence "croakcroak".

This walkthrough simplifies the process of the solution approach, emphasizing the sequence tracking and concurrency accounting that is at the heart of the algorithm.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the function is O(N), where N is the length of the input string croakOfFrogs. This is because the function consists of a single loop that iterates through each character of the input string exactly once. The operations within the loop are constant time lookups in the idx dictionary and simple arithmetic operations, which do not depend on the size of the input string.

Space Complexity

The space complexity of the function is O(1). The additional space required by the algorithm includes a fixed-size cnt array of 5 elements to keep track of the 'croak' sequence counts, the constant size idx dictionary with a mapping from characters to indices, and a few integer variables (ans, x). Since these do not grow with the size of the input, the space required remains constant, irrespective of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.