Minimum Number of Frogs Croaking

Medium

Microsoft

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Topics:

StringsGreedy Algorithms

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.

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.

Constraints:

1 <= croakOfFrogs.length <= 10⁵
croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'.

Try coding it in LeetCode

Minimum Number of Frogs Croaking

Medium

Microsoft

5 views

Topics:

StringsGreedy Algorithms

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.

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.

Constraints:

1 <= croakOfFrogs.length <= 10⁵
croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input string `croakOfFrogs` be empty or null?
What is the maximum length of the input string `croakOfFrogs`?
If the input string is invalid, should I return -1 immediately, or should I process the entire string to ensure no potential for a valid croak exists later?
Are the input characters guaranteed to be only 'c', 'r', 'o', 'a', or 'k'?
If multiple valid croaks are possible with overlapping characters, should I prioritize minimizing the total number of frogs or is there a preferred way to handle overlapping croaks?

Brute Force Solution

Approach

Imagine you're trying to figure out how many frogs need to be singing at the same time to make a given croaking sound. The brute force method is like carefully listening to the sound and figuring out, at each point, the most frogs you could possibly need to make that sound up to that moment.

Here's how the algorithm would work step-by-step:

Listen to the croaking sound one piece at a time, starting from the beginning.
Keep track of how many frogs are potentially in the middle of croaking at each point. A frog is in the middle of croaking if you've heard part of its 'croak' but not the whole thing.
Every time you hear a 'c', imagine a new frog might have started croaking. Increase the number of potentially croaking frogs.
Every time you hear a 'k', imagine a frog has finished its 'croak'. Decrease the number of potentially croaking frogs.
As you listen, keep track of the highest number of frogs that were potentially croaking at the same time. That's the most frogs you ever needed.
Make sure that the croaking sound is actually a valid sequence of 'c', 'r', 'o', 'a', and 'k'. For example, if you hear an 'r' before a 'c', it's an invalid sound.
The highest number of frogs you needed, after checking the sound is valid, is your answer.

Code Implementation

def minimum_number_of_frogs(croak_of_frogs):    maximum_frogs_needed = 0
    current_frogs_croaking = 0
    croak_index = {
        'c': 0,
        'r': 1,
        'o': 2,
        'a': 3,
        'k': 4
    }

    croak_counts = [0] * 5

    for char in croak_of_frogs:
        if char not in croak_index:
            return -1

        index = croak_index[char]
        croak_counts[index] += 1

        # Check for invalid croak sequence
        if index > 0 and croak_counts[index] > croak_counts[index - 1]:
            return -1

        if char == 'c':
            # A new frog starts croaking
            current_frogs_croaking += 1
            maximum_frogs_needed = max(maximum_frogs_needed, current_frogs_croaking)

        elif char == 'k':
            # A frog finishes croaking
            current_frogs_croaking -= 1

    # Check if all frogs finished croaking
    if current_frogs_croaking != 0:
        return -1

    # Verify that all characters appeared the same number of times
    if not all(count == croak_counts[0] for count in croak_counts):
        return -1

    return maximum_frogs_needed

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input string 'croakOfFrogs' of length n once. Inside the loop, it performs constant-time operations such as incrementing/decrementing counters and comparing characters. The number of operations scales linearly with the length of the input string. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm keeps track of the number of potentially croaking frogs using a few counters, one for each character in "croak", plus a variable to store the maximum number of concurrent frogs. These counters and the max_frogs variable require constant space regardless of the length of the input string. Therefore, the auxiliary space used by the algorithm is independent of the input size N and is constant.

Optimal Solution

Approach

The key is to track the progress of each 'croak' independently and reuse frogs as soon as they become available. We're essentially managing a pool of frogs, allocating them to ongoing croaks and freeing them when a croak is complete.

Here's how the algorithm would work step-by-step:

Imagine you have a counter for each letter in the word 'croak'.
As you go through the input string, increase the counter for the corresponding letter.
If you encounter 'c', you either assign a free frog to start a new croak or use a frog that just finished croaking.
For each subsequent letter ('r', 'o', 'a', 'k'), make sure there's a matching previous letter available. If not, the string is invalid.
When you encounter 'k', it means a frog is done croaking, so you make that frog available for a new croak.
Keep track of the maximum number of frogs croaking at the same time, which is the answer.
If at the end, the number of each letter is not the same, the string is invalid.

Code Implementation

def min_number_of_frogs(croak_of_frogs: str) -> int:
    croak_counts = {'c': 0, 'r': 0, 'o': 0, 'a': 0, 'k': 0}
    frogs_in_use = 0
    min_frogs_needed = 0

    for char in croak_of_frogs:
        if char == 'c':
            croak_counts['c'] += 1
            frogs_in_use += 1
            min_frogs_needed = max(min_frogs_needed, frogs_in_use)

        elif char == 'r':
            if croak_counts['c'] <= croak_counts['r']:
                return -1
            croak_counts['r'] += 1

        elif char == 'o':
            if croak_counts['r'] <= croak_counts['o']:
                return -1
            croak_counts['o'] += 1

        elif char == 'a':
            if croak_counts['o'] <= croak_counts['a']:
                return -1
            croak_counts['a'] += 1

        elif char == 'k':
            if croak_counts['a'] <= croak_counts['k']:
                return -1
            croak_counts['k'] += 1

            # Frog is now available to croak again.
            frogs_in_use -= 1

    # Ensure all croaks are complete.
    if not (croak_counts['c'] == croak_counts['r'] == \
            croak_counts['o'] == croak_counts['a'] == croak_counts['k']):
        return -1

    return min_frogs_needed

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input string 'croakOfFrogs' of length n once. Inside the loop, character comparisons and counter updates are performed. Each operation inside the loop takes constant time. Therefore, the overall time complexity is directly proportional to the length of the input string, resulting in O(n) time complexity.

Space Complexity

O(1) – The solution utilizes a fixed number of counters, one for each letter ('c', 'r', 'o', 'a', 'k') in the word 'croak', and potentially a variable to store the maximum number of frogs. The number of these counters and the maximum frog count remain constant regardless of the input string's length (N). Therefore, the auxiliary space used by the algorithm does not scale with the input size, resulting in constant space complexity.

Edge Cases

Case	How to Handle
Null or empty input string	Return -1 immediately as no croaks can be formed.
Input string length is not a multiple of 5	Return -1 because a valid croak must have all 5 characters.
The first character is not 'c'	Return -1 because every croak must start with 'c'.
The order of characters is invalid (e.g., 'craok')	Maintain a count of each character and verify that the order is always c->r->o->a->k, returning -1 if not.
More 'r's than 'c's, or similar imbalances between characters	Return -1 if at any point the count of a character is greater than the count of the character before it in 'croak'.
Input string contains characters other than 'c', 'r', 'o', 'a', 'k'	Return -1 because we only allow characters that form the word croak.
A croak sequence is started but not finished before another croak sequence starts.	Increment frogs when 'c' is encountered and decrement when 'k' is encountered, tracking the maximum concurrent frog count.
After processing the entire string, some croaks are unfinished	Return -1 if the counts of 'c', 'r', 'o', 'a', and 'k' are not equal at the end.