Leetcode 1419: Minimum Number of Frogs Croaking

Input: The input consists of a single string croakOfFrogs that contains characters 'c', 'r', 'o', 'a', and 'k'.

Example: "croakcroak"

Constraints:

• 1 <= croakOfFrogs.length <= 10^5

• The string contains only the characters 'c', 'r', 'o', 'a', 'k'.

Output: The output is an integer representing the minimum number of frogs required to croak all the sounds. If the string is not valid, return -1.

Example: 1

Constraints:

• If the string is valid, return the number of frogs required.

• If the string is not a valid combination, return -1.

Goal: The goal is to count how many frogs are needed to croak all the sounds while ensuring the proper order of letters in each croak.

Steps:

• Count the occurrences of each letter 'c', 'r', 'o', 'a', 'k'.

• Ensure that the letters follow the correct sequence. For example, the count of 'r' cannot exceed the count of 'c', and the same for 'o', 'a', and 'k'.

• Track the maximum number of frogs needed at any point during the croaking process.

Goal: The constraints are mainly focused on the length of the string and the allowed characters.

Steps:

• croakOfFrogs.length can be as large as 100,000.

• The string only contains the characters 'c', 'r', 'o', 'a', and 'k'.

Assumptions:

• The string may contain scrambled sequences of the letters 'c', 'r', 'o', 'a', 'k'.

• A valid croak must follow the exact sequence of letters 'c', 'r', 'o', 'a', and 'k'.

• Input: "croakcroak"

• Explanation: In this example, one frog croaks 'croak' twice, so the minimum number of frogs needed is 1.

• Input: "crcoakroak"

• Explanation: Here, the minimum number of frogs is 2: the first frog croaks 'crcoakroak', and the second frog croaks 'crcoakroak'.

• Input: "croakcrook"

• Explanation: This is an invalid combination because it doesn't form a valid sequence of croaks, so the output is -1.

Approach: To solve this problem, we need to count the occurrences of each letter, ensure they follow the correct sequence, and track the maximum number of frogs needed.

Observations:

• The order of the letters is crucial, so we need to track each frog's croak in sequence.

• We can use counters to ensure that the sequence 'c', 'r', 'o', 'a', 'k' is followed properly and track the maximum number of frogs needed at any point.

Steps:

• Step 1: Initialize counters for each of the five letters in the croak.

• Step 2: Traverse the input string and update the counters based on the order of 'c', 'r', 'o', 'a', 'k'.

• Step 3: Track the maximum number of active frogs needed at any given time.

Empty Inputs:

• An empty string should return -1 because no croaks are made.

Large Inputs:

• Ensure that the solution works efficiently even with strings up to 100,000 characters long.

Special Values:

• Handle cases where characters are out of order, such as 'crak', which should return -1.

Constraints:

• Ensure that the solution adheres to the constraint that the string can only contain 'c', 'r', 'o', 'a', and 'k'.

int minNumberOfFrogs(string croakOfFrogs) {
    int cnt[5] = {}, frogs = 0, max_frog = 0;
    for(auto ch : croakOfFrogs) {
        auto n = string("croak").find(ch);
        ++cnt[n];
        if(n == 0) max_frog = max(max_frog, ++frogs);
        else if(--cnt[n - 1] < 0)   return -1;
        else if(n == 4)             --frogs;
    }
    return frogs == 0? max_frog : -1;
}

1 : Function Definition

int minNumberOfFrogs(string croakOfFrogs) {

Defines the function `minNumberOfFrogs` that takes a string `croakOfFrogs` and returns the minimum number of frogs that could have made those croaks.

2 : Variable Initialization

    int cnt[5] = {}, frogs = 0, max_frog = 0;

Initializes an array `cnt` of size 5 to track the counts of each character ('c', 'r', 'o', 'a', 'k'). Initializes variables `frogs` to track the number of active frogs and `max_frog` to track the maximum number of frogs needed.

3 : Loop Constructs

    for(auto ch : croakOfFrogs) {

Begins a loop that iterates through each character in the input string `croakOfFrogs`.

4 : String Manipulations

        auto n = string("croak").find(ch);

Finds the position of the character `ch` in the string 'croak'. The position corresponds to the order of the letters in the word.

5 : Array Manipulation

        ++cnt[n];

Increments the count for the current character (`ch`) in the `cnt` array.

6 : Conditionals

        if(n == 0) max_frog = max(max_frog, ++frogs);

If the character is 'c' (the first character of 'croak'), it increments the number of frogs and updates the `max_frog` variable to track the maximum number of active frogs.

7 : Conditional Check

        else if(--cnt[n - 1] < 0)   return -1;

If the current character isn't the first one in 'croak', it checks if the previous character count (from `cnt`) is valid. If negative, it indicates an invalid order, so the function returns -1.

8 : Conditional Check

        else if(n == 4)             --frogs;

If the current character is 'k' (the last character in 'croak'), it means a frog has finished its croak, so the frog count is decremented.

9 : Return Statement

    return frogs == 0? max_frog : -1;

At the end of the loop, the function checks if there are any active frogs left. If no frogs are left and the characters followed the proper order, it returns `max_frog` (the maximum number of frogs needed). Otherwise, it returns -1 for an invalid input.

Best Case: O(n), where n is the length of the input string.

Average Case: O(n), as we need to process each character in the string once.

Worst Case: O(n), where n is the length of the input string.

Description: We only need to traverse the input string once to count the letters and verify the sequence.

Best Case: O(1), when the string is valid and processed efficiently.

Worst Case: O(1), since we only need a fixed number of counters (5 for 'c', 'r', 'o', 'a', 'k').

Description: The space complexity is constant, as we only need a fixed amount of space to store the counters for each letter.

Leetcode 1419: Minimum Number of Frogs Croaking

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