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Minimum Number of Frogs Croaking

Number: 1534

Difficulty: Medium

Paid? No

Companies: Roblox, Zoox

Problem Description

Given a string representing interleaved croaks from one or more frogs, where each valid croak consists of the letters "c", "r", "o", "a", "k" in that sequential order, determine the minimum number of frogs needed to produce the given string. If the string cannot be segmented into valid croaks, return -1.

Key Insights

  • Each frog must produce the sequence "c", "r", "o", "a", "k" without reordering.
  • Multiple frogs may be overlapping; one frog might start a new croak as soon as it finishes a previous one.
  • Track the state (or stage) of each frog's current croak to ensure order is maintained.
  • Use counts or an array to store how many frogs are on each character of the croak.
  • As each character is encountered, ensure that it continues a valid sequence from a frog in the expected state or begins a new croak if it is 'c'.
  • At the end, ensure that there are no incomplete croaks.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string since we scan the string once. Space Complexity: O(1) because the extra storage is fixed (only 5 states for the croak).

Solution

The solution uses an array of size 5 to maintain counts for each stage of the croak sequence ("c", "r", "o", "a", and "k"). For each character in the string, the algorithm:

  1. Maps the character to its corresponding position in the sequence.
  2. If the character is 'c', it either reuses a frog that just finished a croak (if available) or increments the count of frogs currently croaking, and updates the maximum number of concurrent frogs.
  3. For a character that is not 'c', it checks if there is an available frog at the previous stage; if yes, it moves that frog to the current stage.
  4. If a valid transition is not possible at any point, the function returns -1.
  5. After processing the entire string, if any croak is incomplete (counts in stages other than 'k' remain), the function will return -1; otherwise, it returns the maximum number of concurrent frogs needed.

Data structures used:

  • An array (or similar structure) with five elements to count frogs at each stage.
  • A mapping from character to its index to quickly check and update states.

This approach ensures that each croak's order is strictly maintained and overlaps are handled optimally.

Code Solutions

# Python solution with detailed comments

def minNumberOfFrogs(croakOfFrogs: str) -> int:
    # The sequence we need to match.
    croak_sequence = "croak"
    # counts[i] will represent the number of frogs at stage i in "croak"
    counts = [0] * len(croak_sequence)
    # To keep track of maximum concurrent frogs required.
    max_frogs = 0

    # Map each character to its corresponding stage index.
    for ch in croakOfFrogs:
        # Find the index corresponding to the character in the croak sequence.
        index = croak_sequence.find(ch)
        if index == -1:
            # Should not happen as input is guaranteed to be one of c, r, o, a, k.
            return -1

        if index == 0:
            # Start of a croak.
            # If possible, reuse a frog that just finished a croak (stage index 4).
            if counts[4]:
                counts[4] -= 1
            else:
                # Otherwise, we need a new frog.
                max_frogs += 1
            counts[0] += 1  # The frog moves to stage 'c'.
        else:
            # For characters other than 'c', check if a frog is at the previous stage.
            if counts[index - 1] == 0:
                # No frog is available to progress, so it's invalid.
                return -1
            # Move a frog from the previous stage.
            counts[index - 1] -= 1
            # If current character is not 'k', update the state to the current stage.
            if index < 4:
                counts[index] += 1
            else:
                # If it's 'k', the frog has finished croaking. We can later reuse this frog.
                counts[4] += 1

    # At the end, valid sequence should not have frogs in intermediate stages.
    if any(counts[i] != 0 for i in range(4)):
        return -1

    return max_frogs

# Example usage:
print(minNumberOfFrogs("croakcroak"))  # Expected output: 1
print(minNumberOfFrogs("crcoakroak"))  # Expected output: 2
print(minNumberOfFrogs("croakcrook"))  # Expected output: -1
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