How do you solve Minimum Number of Frogs Croaking on LeetCode?

Minimum Number of Frogs Croaking (LeetCode #1419) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each character in 'croak' must appear in a specific order; if not, the sequence is invalid.

What is the brute force solution for Minimum Number of Frogs Croaking?

The brute force approach for Minimum Number of Frogs Croaking is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible combination of 'croak' sequences in the string. It counts how many frogs are needed by simulating each possible croak and checking if they can complete their sequence. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Frogs Croaking?

Minimum Number of Frogs Croaking uses the following concepts: String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Frogs Croaking?

Always validate the input for expected patterns before processing. Keep track of the maximum state (like the number of frogs) during iteration to avoid extra passes. Explain your thought process clearly; this shows your understanding of the problem.

What are common mistakes when solving Minimum Number of Frogs Croaking?

Not checking the order of characters properly, leading to invalid sequences. Forgetting to account for the case when the counts of characters are not balanced at the end.

#1419

Minimum Number of Frogs Croaking

Medium

StringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the string while maintaining counts of each character in 'croak'. This allows us to determine how many frogs are croaking simultaneously without needing to check combinations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a count array for 'c', 'r', 'o', 'a', 'k' and a variable for the maximum number of frogs needed.
2Step 2: Iterate through each character in the string and update the count based on the character.
3Step 3: For each character, ensure that the previous character's count is not greater than the current character's count.
4Step 4: Update the maximum number of frogs needed when encountering 'c'.
5Step 5: At the end, check if the counts are balanced and return the maximum number of frogs or -1 if invalid.

solution.py17 lines

1def minNumberOfFrogs(croakOfFrogs):
2    croak_count = [0] * 5
3    frogs = 0
4    max_frogs = 0
5    croak = 'croak'
6    for char in croakOfFrogs:
7        if char not in croak:
8            return -1
9        croak_count[croak.index(char)] += 1
10        if char == 'c':
11            frogs += 1
12            max_frogs = max(max_frogs, frogs)
13        if char == 'k':
14            frogs -= 1
15        if any(crook_count[i] < croak_count[i-1] for i in range(1, 5)):
16            return -1
17    return max_frogs if all(count == 0 for count in croak_count) else -1

ℹ

Complexity note: The time complexity is O(n) since we only make a single pass through the string, updating counts as we go. The space complexity is O(1) because we only use a fixed amount of space for the counts.

1Each character in 'croak' must appear in a specific order; if not, the sequence is invalid.
2The maximum number of frogs at any point corresponds to the number of 'c's encountered minus the number of 'k's.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.