What is the time complexity of Minimum Number of Frogs Croaking?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Number of Frogs Croaking?

This problem uses String, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Frogs Croaking?

Insight 1: The order of characters 'c', 'r', 'o', 'a', 'k' is crucial. A character can only appear if the preceding character in "croak" has appeared at least as many times. Insight 2: We can maintain counts of each character and check if the order is maintained. If not, the string is invalid.

Which companies ask LeetCode 1419. Minimum Number of Frogs Croaking in interviews?

This problem is frequently asked at Roblox, Zoox.

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Minimum Number of Frogs Croaking - LeetCode 1419 Solution

Minimum Number of Frogs Croaking - Complete Solution Guide

Minimum Number of Frogs Croaking is LeetCode problem 1419, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You 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 'c' , 'r' , 'o' , 'a' , and 'k' sequentially . The frogs have to print all five letters to finish a croak. If the given string is not a combination of a vali

Detailed Explanation

The problem asks us to determine the minimum number of frogs required to produce a given string `croakOfFrogs`, where the string is a combination of multiple "croak" sequences from different frogs possibly croaking at the same time. A frog must complete the entire "croak" sequence before it can croak again. If the input string is not a valid combination of "croak" sequences, we should return -1. The constraints are the length of the string is between 1 and 10^5, and the string contains only 'c', 'r', 'o', 'a', and 'k'.

Solution Approach

The solution iterates through the given string `croakOfFrogs`. During each iteration, the algorithm updates the counts of characters 'c', 'r', 'o', 'a', and 'k'. It verifies if the counts are in the correct order ('c' >= 'r' >= 'o' >= 'a' >= 'k'). If the order is ever violated, it signifies an invalid 'croak' sequence, and the function returns -1. The algorithm also calculates the number of active frogs (frogs that started croaking but haven't finished) by subtracting the count of 'k' from the count of 'c'. The maximum number of active frogs encountered during the iteration is tracked. Finally, the algorithm checks if the count of all the characters is the same, meaning every frog has finished croaking, and returns the maximum number of active frogs. If the counts are not the same, the string is invalid and -1 is returned.

Step-by-Step Algorithm

Step 1: Initialize counters for 'c', 'r', 'o', 'a', 'k' to 0.
Step 2: Initialize a variable `max_frogs` to 0. This will store the maximum number of active frogs at any point.
Step 3: Iterate through the input string `croakOfFrogs` character by character.
Step 4: For each character, increment the corresponding counter.
Step 5: Check if the order of counts is maintained (c >= r >= o >= a >= k). If not, return -1.
Step 6: Calculate the number of active frogs as `c_count - k_count`.
Step 7: Update `max_frogs` with the maximum value between the current `max_frogs` and the number of active frogs.
Step 8: After iterating through the entire string, check if all the counts are equal. If they are, return `max_frogs`; otherwise, return -1.

Key Insights

Insight 1: The order of characters 'c', 'r', 'o', 'a', 'k' is crucial. A character can only appear if the preceding character in "croak" has appeared at least as many times.
Insight 2: We can maintain counts of each character and check if the order is maintained. If not, the string is invalid.
Insight 3: The number of active frogs at any point is the difference between the number of 'c' and 'k' encountered so far. We need to keep track of the maximum number of active frogs during the iteration.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Counting.

Companies

Asked at: Roblox, Zoox.

Minimum Number of Frogs Croaking | LeetCode 1419

Medium

String Counting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int c_count = 0, r_count = 0, o_count = 0, a_count = 0, k_count = 0;
        int max_frogs = 0;

        for (int i = 0; i < croakOfFrogs.length(); i++) {
            char charAt = croakOfFrogs.charAt(i);
            if (charAt == 'c') {
                c_count += 1;
            } else if (charAt == 'r') {
                r_count += 1;
            } else if (charAt == 'o') {
                o_count += 1;
            } else if (charAt == 'a') {
                a_count += 1;
            } else if (charAt == 'k') {
                k_count += 1;
            }

            if (!(c_count >= r_count && r_count >= o_count && o_count >= a_count && a_count >= k_count)) {
                return -1;
            }

            int active_frogs = c_count - k_count;
            max_frogs = Math.max(max_frogs, active_frogs);
        }

        if (c_count == r_count && r_count == o_count && o_count == a_count && a_count == k_count) {
            // A valid non-empty string must have c_count > 0.
            // If all counts are equal, this implies it's a valid combination.
            return max_frogs;
        } else {
            return -1;
        }
    }
}

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Problem Info

Description

Problem 1419: Minimum Number of Frogs Croaking - Detailed explanation and solution approaches.

Topics

String Counting

Companies

RobloxZoox

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1419

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize counters for 'c', 'r', 'o', 'a', 'k' to 0.
2
Step 2: Initialize a variable `max_frogs` to 0. This will store the maximum number of active frogs at any point.
3
Step 3: Iterate through the input string `croakOfFrogs` character by character.
4
Step 4: For each character, increment the corresponding counter.
5
Step 5: Check if the order of counts is maintained (c >= r >= o >= a >= k). If not, return -1.
6
Step 6: Calculate the number of active frogs as `c_count - k_count`.
7
Step 7: Update `max_frogs` with the maximum value between the current `max_frogs` and the number of active frogs.
8
Step 8: After iterating through the entire string, check if all the counts are equal. If they are, return `max_frogs`; otherwise, return -1.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single loop that iterates through the input string. The space complexity is constant because only a fixed number of integer variables are used to store the counts and the maximum number of frogs.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Stack-based approach

We could potentially use stacks. For each 'c', we push a frog onto a stack. When we see an 'r', we pop a frog and assign it to the 'r'. Continue until 'k', then a frog is released. This is harder to implement and less efficient.

Time: O(n)Space: O(n) in the worst case where the input is all 'c's

Trade-offs: The stack-based approach is more complex to implement and has a higher space complexity than the counting approach. The counting approach is preferred for its simplicity and efficiency.

✨

Key Insights

1Insight 1: The order of characters 'c', 'r', 'o', 'a', 'k' is crucial. A character can only appear if the preceding character in "croak" has appeared at least as many times.
2Insight 2: We can maintain counts of each character and check if the order is maintained. If not, the string is invalid.
3Insight 3: The number of active frogs at any point is the difference between the number of 'c' and 'k' encountered so far. We need to keep track of the maximum number of active frogs during the iteration.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The provided solution implicitly handles an empty string correctly because the counts remain 0, which are equal and the max_frogs also will be 0 and gets returned.
Edge case 2: Input contains characters other than 'c', 'r', 'o', 'a', 'k'. The given problem description doesn't specify how to handle cases with characters besides 'c', 'r', 'o', 'a', 'k'. However, the code only increments the counters for these 5 characters, implicitly ignoring other characters.
Edge case 3: String like "croakcrook". The example shows such case should return -1. The solution handles it correctly by checking the condition c_count >= r_count >= o_count >= a_count >= k_count

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the order of characters is maintained. If the order is not maintained (e.g., 'r' appears before 'c'), then the string is invalid.

✗Common mistake 2: Not checking if all the counts are equal at the end. If the counts are not equal, it means some frogs didn't finish croaking, and the string is invalid.

✗Common mistake 3: Incorrectly calculating the number of active frogs. The number of active frogs is the number of frogs that have started croaking ('c') but haven't finished ('k'). So active_frogs = c_count - k_count.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine a system where resources are represented by 'c', 'r', 'o', 'a', 'k'. The algorithm can be used to determine the minimum number of resource sets required to process a given sequence of resource requests.

🏦Real-world scenario 2: Log analysis. Analyzing logs where specific events need to happen in a specific order. You could use this kind of algorithm to ensure all steps happen and calculate the minimum number of entities completing the process

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Number of Frogs Croaking"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether counting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?