Minimum Number Of Frogs Croaking

Problem Statement:

We are given the string croakOfFrogs, which represents a combination of the string "croak" from different frogs, that is, multiple frogs can croak at the same time, so multiple "croak" are mixed.

Return the minimum number of different frogs to finish all the croaks in the given string.

A valid "croak" means a frog is printing five letters sequentially: 'c', 'r', 'o', 'a', and 'k'. The frogs have to print all five letters to finish a croak. If the given string is not a combination of a valid "croak," return -1.

Java Implementation

Java Approach 1

import java. util.Scanner; 
public class FrogCroak {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in); 
        // Input: Croak of frogs string
System.out.print("Enter the croak of frogs string: ");
        String croakOfFrogs = scanner.nextLine();
        Solution solution = new Solution();
        // Output: Minimum number of frogs after applying minNumberOfFrogs method
        int result = solution.minNumberOfFrogs(croakOfFrogs); 
System.out.println("Result: " + result); 
scanner.close();
    }
}
class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int c = 0, r = 0, o = 0, a = 0;
        int max = 0; 
        // Iterate through each character in the croakOfFrogs string
        for (char ch: croakOfFrogs.toCharArray()) {
                //if condition if ch equals to ?c we have to increase the c by 1
            if (ch == 'c') {
                c++;
            } 
              // else if the character is ?r? we will go inside the loop and follow conditional statements 
           else if (ch == 'r') {
                // Check if 'r' comes after 'c'
                if (c < ++r)   
return -1;
            } else if (ch == 'o') {
                // Check if 'o' comes after 'r'
                if (r < ++o)       
return -1;
            } else if (ch == 'a') {
                // Check if 'a' comes after 'o'
                if (o < ++a) 
return -1;
            } else {
                // Check if 'k' comes after 'a' and update counts
                max = Math.max(max, c--);
                o--;
                r--;
                a--;
            }
        }
        // Check if counts are zero to ensure valid croak
        return (c == 0 && r == 0 && o == 0 && a == 0) ? max: -1;
    }
}

Output:

Code Explanation:

The code defines a Solution class with a minNumberOfFrogs method, which takes a string representing the croak of frogs. It iterates through each character, incrementing counts for 'c,' 'r,' 'o', and 'a'. If the order of appearance is violated, it returns -1.
For 'k,' it updates the maximum count of frogs, decrements count and checks the validity of the croak. The method returns the minimum number of frogs required for the valid croak.

Time Complexity:

The time complexity is O(N), N being the size of input string spellingOfFrogs. The algorithm do a single run over the entire string by iterating the operation which involves a constant time.

Space Complexity:

Though space complexity is O(1). To show this, the method contacts the constant extra space allocation of the tally value r and c. The space complexity doesn't depend on the size of the input.

Drawback:

The weakness of the solution is that it involves counting the frequency of each letter sequentially and assumes certain versioning. In case the input does not obey this order or is incomplete croak, the solution may be wrong.

Java Approach 2

import java. util.HashMap;
import java. util.Map;
import java. util.Scanner; 
public class FrogMinCounter {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in); 
        // Input the croakOfFrogs string
System.out.print("Enter the croakOfFrogs string: ");
        String croakOfFrogs = scanner.nextLine();
        Solution solution = new Solution();
        int result = solution.minNumberOfFrogs(croakOfFrogs); 
        // Output the result
System.out.println("Result: " + result); 
scanner.close();
    }
}
class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int n = croakOfFrogs.length();
        int idx = 0;
        int ans = 0;
        int curr = 0;
        int max_frogs = n / 5; 
        // Check if the length of the string is a multiple of 5
        if (n % 5 != 0) return -1; 
        int[] freq = new int[]{0, 0, 0, 0, 0};
Map<Character, Integer> map = new HashMap<>();
     //fill the hashmap with the characters 
   map.put('c', 0);
        map.put('r', 1);
        map.put('o', 2);
        map.put('a', 3);
        map.put('k', 4); 
        while (idx < croakOfFrogs.length()) {
            Character ch = croakOfFrogs.charAt(idx);
            // Check if the character is valid
            if (!map.containsKey(ch)) return -1; 
                        //if condition if ch equals to 'c we have to increase the c by 1
    if (ch == 'k') {
curr--;
            } else {
freq[map.get(ch)]++;
                if (ch == 'c') {       
curr++;           
ans = Math.max(ans, curr);
                } else {       
freq[map.get(ch) - 1]--;
                    // Check if the frequency becomes negative
                    if (freq[map.get(ch) - 1] == -1) return -1;
                }
            }
            idx++;
        }
        // Check if the 'c' frequency is zero
        if (freq[0] > 0) return -1;
        return ans;
    }
}

Output:

Code Explanation:

The code defines a class Solution with a method minNumberOfFrogs to find the minimum number of frogs needed to produce a given croak sequence. The algorithm uses an array to store the frequencies of each croak sound ('c,' 'r,' 'o', 'a', 'k') and iterates through the input string. It checks if the sequence is valid by incrementing and decrementing counts based on croak occurrences. The maximum count of 'c' at any point represents the minimum number of frogs required. The program also ensures the length of the input is a multiple of 5 and returns -1 if any invalid condition is encountered.

Time Complexity:

The time complexity comes out to be O(n), for each n value represent length of an input string croakOfFrogs. This is a core operation which is done on each character as well as traversing the string once.

Space Complexity:

The space complexity is O(1) due to the constant memory consumption for freq array and map. It is size-independent.

Drawback:

The proposed solution's weakness is its dependence on some specific character indices and frequency arrays, which makes it less flexible regarding changes taking place with respect to the input or modifications made to the croak pattern.
The method assumes a particular order of characters ('c', 'r', 'o' , 'a', and k) that requires additional negative frequency checks. This strict framework may result in errors as the croakOfFrogs string does not fit into a given format.

Java Approach 3

import java. util.Scanner; 
public class FrogCounter {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in); 
        // Input the croakOfFrogs string
System.out.print("Enter the croakOfFrogs string: ");
        String croakOfFrogs = scanner.next();
        Solution solution = new Solution();
        int result = solution.minNumberOfFrogs(croakOfFrogs); 
        // Output the result
System.out.println("Number of frogs: " + result);
        scanner.close();
    }
}
class Solution {
    public int getIndex(char c) {
        // Return the index based on the character
        switch (c) {
            case 'c':
                return 0;
            case 'r':
                return 1;
            case 'o':
                return 2;
            case 'a':
                return 3;
            case 'k':
                return 4;
        }
        return -1;
    }
    public int minNumberOfFrogs(String croakOfFrogs) {
        int croak[] = new int[5]; // c->0,r->1,.....k->4
        char ar[] = croakOfFrogs.toCharArray();
        int n = ar.length;
        int frogs = 0; 
        for (int i = 0; i < n; i++) {
            int index = getIndex(ar[i]);
            croak[index]++; 
            // Cases like "crroak" where curr character is greater than the prev character - these are not valid strings
            if (index > 0 && croak[index] > croak[index - 1])
                return -1; 
            // Counting the number of croaks which are currently started, which will be equal to the number of 'c'
            frogs = Math.max(frogs, croak[0]); 
            if (index == 4) {
                // If croak is complete, release all frogs
                for (int j = 0; j <= 4; j++)
croak[j]--;
            }
        }
        // For cases like "ccrroa," which is not complete "croak."
        for (int i = 0; i < 4; i++)
            if (croak[i] > 0)
                return -1; 
        if (frogs == 0)
            frogs = -1; 
        return frogs;
    }
}

Output:

Code Explanation:

The code defines a class Solution with a method minNumberOfFrogs that calculates the minimum number of frogs needed to produce a given sequence of frog sounds. It uses an array croak to keep track of the count of each sound ('c', 'r', 'o', 'a', 'k').
The function iterates through the input string, updating the count of each sound and checking for the validity of the sequence. The index mapping function getIndex ensures proper indexing. The result is the maximum count of the 'c' sound, representing the minimum number of frogs required.

Time Complexity:

The running time of minNumberOfFrogs is O(n) and n is defined as the part of CroakOfFrogs that we need to determine. This function does not move through the string, but it deals with each character separately. Every time, the function performs a constant-time operation for each character.

Space Complexity:

The space complexity of the process being managed is O(1), because the auxiliary storage like croak array and index variable required is considered to be constant. Regarding to space complexity, it doesn't depend on how long the string is.

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