Minimum Number Of Frogs Croaking - Leetcode Solution

The Minimum Number of Frogs Croaking problem on LeetCode is a medium-level problem that involves string manipulation. The problem statement is as follows:

You have a string s that represents a sequence of frogs croaking. Each 'croak' of a frog is represented by a character in the string. The four characters are 'c', 'r', 'o', and 'a'. You need to determine the minimum number of frogs that are needed to complete the sequence. A valid sequence is one where the frogs croak in the order 'croak'. In other words, for a valid sequence, the first 'c' must be followed by an 'r', then an 'o', and finally an 'a'. Once a frog has croaked 'a', it is considered done and can no longer contribute to the sequence.

To solve this problem, we can use a hashmap to keep track of the number of frogs that are currently croaking and a variable to keep track of the maximum number of frogs that have been croaking at the same time. We iterate through the string s and for each character we check if it is a valid frog character. If it is, we update the hashmap accordingly. If at any point the sequence is not valid, we return -1. Finally, we return the maximum number of frogs that have been croaking at the same time.

Here is an implementation of the solution in Python:

def minNumberOfFrogs(croakOfFrogs: str) -> int:
    count = {"c": 0, "r": 0, "o": 0, "a": 0}
    max_count = 0
    for c in croakOfFrogs:
        if c not in count:
            return -1
        if c == "c":
            count["c"] += 1
        elif c == "r":
            if count["c"] <= count["r"]:
                return -1
            count["r"] += 1
        elif c == "o":
            if count["r"] <= count["o"]:
                return -1
            count["o"] += 1
        elif c == "a":
            if count["o"] <= count["a"]:
                return -1
            count["a"] += 1
            count["c"] -= 1
            count["r"] -= 1
            count["o"] -= 1
        max_count = max(max_count, count["c"])
    if sum(count.values()) > 0:
        return -1
    return max_count

The time complexity of this solution is O(n), where n is the length of the string s. This is because we only iterate through the string once and perform constant time operations for each character. The space complexity is also O(1), as the size of the hashmap is constant and independent of the length of the string.

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