#1419

Medium

LeetCode Problem Workspace

Minimum Number of Frogs Croaking

Determine the minimum number of frogs required to sequentially produce all croaks in a mixed frog croak string efficiently.

string counting

Problem Statement

You are given a string croakOfFrogs representing overlapping frog croaks, where multiple frogs can croak simultaneously and their sequences may interleave. Each frog must complete the full sequence 'c','r','o','a','k' to finish a croak.

Return the minimum number of frogs required to produce all croaks in croakOfFrogs. If the string contains invalid sequences that cannot form complete 'croak' patterns, return -1.

Examples

Example 1

Input: croakOfFrogs = "croakcroak"

Output: 1

One frog yelling "croak" twice.

Example 2

Input: croakOfFrogs = "crcoakroak"

Output: 2

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

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'.

Solution Approach

Sequential Counting with Frequency Map

Use a hashmap to track counts of each letter in 'croak'. For every character, increment its count and check that no preceding letter has a lower count. This ensures all partial croaks maintain correct ordering.

Track Active Frogs

Maintain a counter for frogs currently croaking. Increment when a 'c' starts and decrement when a 'k' completes. The peak value represents the minimum number of frogs needed.

Validate Complete Sequences

After processing the string, ensure all letter counts are balanced so every 'c' has a matching 'k'. Any mismatch indicates an invalid input, returning -1 immediately.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) because we iterate through the string once, updating counts and checking order. Space complexity is O(1) since the hashmap only stores fixed counts for 'c','r','o','a','k'.

What Interviewers Usually Probe

Check for simultaneous overlapping croaks to assess string plus counting comprehension.
Look for proper validation of sequence order to detect partial or invalid croaks.
Ensure space usage is constant, not scaling with string length, demonstrating counting efficiency.

Common Pitfalls or Variants

Common pitfalls

Failing to track multiple frogs simultaneously, leading to undercounting.
Not validating order, allowing sequences like 'crook' to pass incorrectly.
Assuming counts reset after each 'k', missing overlapping croaks from other frogs.

Follow-up variants

Input with multiple overlapping croaks requiring dynamic frog tracking.
Strings where some croak letters are missing or out of order to test validation.
Extended patterns with repeated croak sequences increasing concurrency requirements.

FAQ

What is the main challenge in Minimum Number of Frogs Croaking?

The challenge is correctly counting overlapping 'croak' sequences and tracking simultaneous frogs without missing partial sequences.

Can I solve this problem with simple string replacement?

No, string replacement fails when croaks overlap; you must use counting and order tracking for accuracy.

How do I know if the input string is invalid?

Check if any letter in 'croak' appears out of sequence or if total counts of 'c','r','o','a','k' are not balanced at the end.

Is there a limit to the number of frogs the string can represent?

Yes, the maximum simultaneous frogs is determined by the peak number of active croaks while processing the string.

Which pattern does this problem exemplify?

It follows the String plus Counting pattern, focusing on sequential character tracking and frequency management to find concurrency.

terminal

Solution

Solution 1: Counting + Simulation

We note that if the string `croakOfFrogs` is composed of several valid `"croak"` characters mixed together, its length must be a multiple of $5$. Therefore, if the length of the string is not a multiple of $5$, we can directly return $-1$.

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class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        if len(croakOfFrogs) % 5 != 0:
            return -1
        idx = {c: i for i, c in enumerate('croak')}
        cnt = [0] * 5
        ans = x = 0
        for i in map(idx.get, croakOfFrogs):
            cnt[i] += 1
            if i == 0:
                x += 1
                ans = max(ans, x)
            else:
                if cnt[i - 1] == 0:
                    return -1
                cnt[i - 1] -= 1
                if i == 4:
                    x -= 1
        return -1 if x else ans

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