Introduction

In this blog post, we will explore how to solve the LeetCode problem "1419" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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'.

Explanation

Here's a breakdown of the solution:

• Track Frog Progress: Maintain a count of frogs at each stage of the "croak" sequence ('c', 'r', 'o', 'a', 'k').
• Reuse Frogs: When a 'c' is encountered, try to reuse a frog that has finished croaking ('k'). If none are available, start a new frog.

• Validate and Count: Ensure that the sequence is valid (e.g., 'r' doesn't appear before 'c'). The maximum number of active frogs at any point is the answer.

• Runtime Complexity: O(n), where n is the length of the input string. Storage Complexity: O(1) - constant extra space.

Code

    def min_number_of_frogs(croakOfFrogs: str) -> int:
    """
    Calculates the minimum number of frogs required to produce the given croaking sequence.

    Args:
        croakOfFrogs: A string representing the combined croaking sequence from multiple frogs.

    Returns:
        The minimum number of frogs needed, or -1 if the sequence is invalid.
    """

    counts = {'c': 0, 'r': 0, 'o': 0, 'a': 0, 'k': 0}
    frogs = 0  # Current number of active frogs
    max_frogs = 0  # Maximum number of active frogs seen so far

    for char in croakOfFrogs:
        if char == 'c':
            if counts['k'] > 0:
                counts['k'] -= 1
            else:
                frogs += 1
            counts['c'] += 1
        elif char == 'r':
            if counts['c'] == 0:
                return -1
            counts['c'] -= 1
            counts['r'] += 1
        elif char == 'o':
            if counts['r'] == 0:
                return -1
            counts['r'] -= 1
            counts['o'] += 1
        elif char == 'a':
            if counts['o'] == 0:
                return -1
            counts['o'] -= 1
            counts['a'] += 1
        elif char == 'k':
            if counts['a'] == 0:
                return -1
            counts['a'] -= 1
            counts['k'] += 1
            frogs -=1
        else:
            return -1

        max_frogs = max(max_frogs, frogs)

    if counts['c'] != 0 or counts['r'] != 0 or counts['o'] != 0 or counts['a'] != 0:
        return -1

    return max_frogs