LeetCode 1419 - Minimum Number of Frogs Croaking

LeetCode Problem 1419

Difficulty: 🟡 Medium
Topics: String, Counting

Solution

Problem Understanding

The problem asks us to determine the minimum number of frogs needed to produce a given sequence of croaks represented by the string croakOfFrogs. Each frog croaks in the sequence "c" -> "r" -> "o" -> "a" -> "k", and multiple frogs can croak simultaneously. The challenge is that the letters from different frogs can interleave, but each frog must complete the full "croak" in order. The input string only contains characters 'c', 'r', 'o', 'a', 'k', and the length can be up to 100,000 characters.

The expected output is an integer representing the minimum number of frogs required to complete all croaks concurrently, or -1 if the string is not a valid combination of "croak" sequences. Edge cases to consider include incomplete croaks, overlapping croaks where letter order is violated, and scenarios where multiple frogs are croaking in parallel.

The constraints tell us that the solution needs to be efficient (O(n) or close) because of the potential size of the input. The problem guarantees that only valid croak characters are present, but not that they form valid croaks.

Approaches

A brute-force approach would attempt to simulate every possible frog croaking sequence by assigning letters to different frogs dynamically. For each character, we would try to continue an existing frog's croak or start a new one. This is cumbersome, slow, and difficult to implement correctly for large inputs, as it could approach O(n^2) time complexity when checking which frog can take the next character.

The key insight for an optimal approach is to track the count of frogs in each stage of the croak sequence using a fixed-size array or hash map. Specifically, we maintain counts for how many frogs are currently at 'c', 'r', 'o', 'a', and 'k'. Each new character either advances a frog to the next stage or starts a new frog at 'c'. By keeping track of active frogs at each stage, we can determine the maximum number of simultaneous croaks, which gives the minimum number of frogs needed. If at any point the counts are inconsistent (e.g., more 'r' than 'c'), the input is invalid and we return -1.

Algorithm Walkthrough

1. Initialize a fixed array counts of size 5 to track the number of frogs at each stage of "c", "r", "o", "a", "k".
2. Initialize a variable max_frogs to track the maximum number of simultaneous frogs.
3. Iterate through each character in croakOfFrogs.
4. Map each character to its stage index: 'c'->0, 'r'->1, 'o'->2, 'a'->3, 'k'->4.
5. If the character is 'c', increment counts[0] since it starts a new croak.
6. For other characters, check if the previous stage has a positive count. If so, decrement the previous stage and increment the current stage. If not, return -1 because the croak sequence is broken.
7. After updating counts, compute the number of frogs currently croaking as sum(counts[:4]) (exclude 'k' as completed croaks free up frogs). Update max_frogs if this number is higher.
8. After iterating through the string, check that all counts except 'k' are zero. If not, return -1.
9. Return max_frogs.

Why it works: By maintaining counts at each stage, we ensure the order of "croak" is preserved, and by tracking the maximum concurrent frogs, we get the minimum number required. The algorithm guarantees correctness because any violation of the croak order is immediately caught.

Python Solution

class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        stages = {'c': 0, 'r': 1, 'o': 2, 'a': 3, 'k': 4}
        counts = [0] * 5
        max_frogs = 0

        for char in croakOfFrogs:
            index = stages[char]
            if index == 0:
                counts[0] += 1
            else:
                if counts[index - 1] == 0:
                    return -1
                counts[index - 1] -= 1
                counts[index] += 1

            current_frogs = sum(counts[:4])
            max_frogs = max(max_frogs, current_frogs)

        if any(counts[:4]):
            return -1
        return max_frogs

This implementation uses a fixed array counts to track the number of frogs at each croak stage. Each character either advances a frog or starts a new one. The maximum number of frogs simultaneously in progress is tracked by summing the counts of stages 'c' through 'a'. The function validates the input by ensuring that all croaks are completed correctly.

Go Solution

func minNumberOfFrogs(croakOfFrogs string) int {
    stages := map[rune]int{'c': 0, 'r': 1, 'o': 2, 'a': 3, 'k': 4}
    counts := [5]int{}
    maxFrogs := 0

    for _, char := range croakOfFrogs {
        index := stages[char]
        if index == 0 {
            counts[0]++
        } else {
            if counts[index-1] == 0 {
                return -1
            }
            counts[index-1]--
            counts[index]++
        }

        currentFrogs := counts[0] + counts[1] + counts[2] + counts[3]
        if currentFrogs > maxFrogs {
            maxFrogs = currentFrogs
        }
    }

    for i := 0; i < 4; i++ {
        if counts[i] != 0 {
            return -1
        }
    }
    return maxFrogs
}

The Go implementation closely mirrors the Python version. Differences include using rune for characters and a fixed-size array for counts. Index mapping uses a map to convert characters to stage indices. All edge cases are handled identically.

Worked Examples

Example 1: "croakcroak"

Result: 1

Example 2: "crcoakroak"