1419. Minimum Number of Frogs Croaking

Problem Description

Given a string croakOfFrogs representing a combination of "croak" sounds from different frogs, determine the minimum number of different frogs needed to produce the given string. A valid "croak" is a sequence of 'c', 'r', 'o', 'a', 'k'. Frogs must croak sequentially, and if the input string is not a valid combination of "croak"s, return -1.

Solution Approach: Counting and Simulation

This problem can be efficiently solved using a counting approach combined with simulation. The core idea is to track the number of frogs at each stage of the croaking process ('c', 'r', 'o', 'a', 'k').

1. Validity Check: If the length of croakOfFrogs is not divisible by 5 (the length of "croak"), it's impossible to form valid "croaks", so return -1.

2. Counting Array: We'll use an array cnt of size 5 to store the counts of each character ('c', 'r', 'o', 'a', 'k'). cnt[i] represents the number of frogs currently at stage i of their croak (0 for 'c', 1 for 'r', etc.).

3. Active Frogs: A variable x will track the number of currently active frogs (those that have started but not finished their croak). The maximum value of x during the simulation represents the minimum number of frogs needed.

4. Iteration: Iterate through the input string:

   • If the character is 'c', increment x (a new frog starts) and update the maximum number of active frogs.
   • If the character is 'r', 'o', 'a', or 'k', check if there are enough frogs in the previous stage (cnt[i-1] > 0). If not, return -1 (invalid sequence). Decrement the count of the previous stage and, if it's 'k', decrement x (a frog finishes its croak).

5. Final Check: After the iteration, if x is 0, all frogs have finished; return the maximum number of active frogs. Otherwise, return -1 (incomplete croaks).

Time and Space Complexity

• Time Complexity: O(n), where n is the length of croakOfFrogs. We iterate through the string once.
• Space Complexity: O(1). We use a fixed-size array cnt and a few integer variables.

Code Implementation

Here's the code in several programming languages:

Python

class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        n = len(croakOfFrogs)
        if n % 5 != 0:
            return -1
        cnt = [0] * 5
        ans = x = 0
        croak_map = {'c': 0, 'r': 1, 'o': 2, 'a': 3, 'k': 4}
        for c in croakOfFrogs:
            i = croak_map[c]
            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 ans if x == 0 else -1
 

Java

class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int n = croakOfFrogs.length();
        if (n % 5 != 0) return -1;
        int[] cnt = new int[5];
        int ans = 0, x = 0;
        for (char c : croakOfFrogs.toCharArray()) {
            int i = "croak".indexOf(c);
            cnt[i]++;
            if (i == 0) {
                x++;
                ans = Math.max(ans, x);
            } else {
                if (cnt[i - 1] == 0) return -1;
                cnt[i - 1]--;
                if (i == 4) x--;
            }
        }
        return x == 0 ? ans : -1;
    }
}

C++

class Solution {
public:
    int minNumberOfFrogs(string croakOfFrogs) {
        int n = croakOfFrogs.length();
        if (n % 5 != 0) return -1;
        vector<int> cnt(5, 0);
        int ans = 0, x = 0;
        for (char c : croakOfFrogs) {
            int i = "croak".find(c);
            cnt[i]++;
            if (i == 0) {
                x++;
                ans = max(ans, x);
            } else {
                if (cnt[i - 1] == 0) return -1;
                cnt[i - 1]--;
                if (i == 4) x--;
            }
        }
        return x == 0 ? ans : -1;
    }
};

These implementations follow the algorithm described above, providing a concise and efficient solution to the problem. Remember to adapt the code to your preferred programming language and environment.