Minimum Number of Frogs Croaking - Practice Coding Problems

Minimum Number of Frogs Croaking - Problem

String Counting Medium

🐸 Picture a peaceful pond where multiple frogs are croaking simultaneously! You're given a string croakOfFrogs that represents the overlapping sounds of different frogs, where each frog must say "croak" in the exact sequence: 'c' → 'r' → 'o' → 'a' → 'k'.

Your task is to determine the minimum number of frogs needed to produce all the croaking sounds in the given string. Each frog can only say one letter at a time and must complete the full sequence "croak" before starting again.

Important: If the input string cannot be formed by valid frog croaking (e.g., letters out of order, incomplete sequences), return -1.

Example: In the string "crcoroakoak", we need at least 2 frogs because they overlap in their croaking patterns.

Input & Output

example_1.py — Basic Valid Case

$ Input: croakOfFrogs = "croakcroak"

› Output: 1

💡 Note: One frog can say the entire sequence: first "croak", then another "croak". Since there's no overlap, we only need 1 frog.

example_2.py — Overlapping Croaks

$ Input: croakOfFrogs = "crcoroakoak"

› Output: 2

💡 Note: Two frogs are needed: Frog1 says "cr" then "oak", while Frog2 says "co" then "roak". They overlap in the middle, so we need 2 frogs minimum.

example_3.py — Invalid Sequence

$ Input: croakOfFrogs = "croakcrook"

› Output: -1

💡 Note: Invalid because we have 'croakcrook' where the second sequence has 'crook' instead of 'croak'. The 'k' appears before 'a', violating the sequence.

Constraints

1 ≤ croakOfFrogs.length ≤ 10⁵
croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'
Each frog must complete the exact sequence 'c' → 'r' → 'o' → 'a' → 'k'

Visualization

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Understanding the Visualization

Start State Machine

Initialize counters for each possible frog state

Process Characters

For each character, move frogs to next state or start new frogs

Track Peak Usage

Monitor the maximum number of simultaneous active frogs

Validate Completion

Ensure all frogs complete their full 'croak' sequence

Key Takeaway

🎯 Key Insight: By tracking frog states with counters, we avoid complex simulations and solve in optimal O(n) time. The peak concurrent frogs during the process is our answer!

Asked in

G Google 45 a Amazon 35 ⊞ Microsoft 25 f Meta 20

The optimal solution uses a state counter approach to track frogs at each stage of saying "croak". We maintain counters for frogs at states 'c', 'cr', 'cro', 'croa' and move them through states as we process each character. The maximum concurrent active frogs during this process gives us our answer in O(n) time and O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Assignments)	O(5^n)	O(n)	Try assigning each character to different frogs and check all valid combinations
State Counter (Optimal)	O(n)	O(1)	Track frog states (c, cr, cro, croa) with counters in single pass

Brute Force (Try All Assignments) — Algorithm Steps

Generate all possible ways to group characters into frog sequences
For each grouping, check if every frog follows the 'c'→'r'→'o'→'a'→'k' pattern
Track the minimum number of frogs across all valid groupings
Return -1 if no valid grouping exists

Visualization

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Step-by-Step Walkthrough

Start with first character

Try assigning first 'c' to different frogs

Branch for each assignment

For each frog assignment, try next character

Validate complete paths

Check if assignment creates valid 'croak' sequences

Return minimum

Find the path with minimum number of frogs

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>

int isValidAssignment(char* croak, int* assignment, int len) {
    int frogStates[1000] = {0}; // Assume max 1000 frogs
    int frogExists[1000] = {0};
    char* expected = "croak";
    
    for (int i = 0; i < len; i++) {
        int frogId = assignment[i];
        char ch = croak[i];
        
        if (!frogExists[frogId]) {
            frogExists[frogId] = 1;
            frogStates[frogId] = 0;
        }
        
        if (expected[frogStates[frogId]] != ch) {
            return 0;
        }
        
        frogStates[frogId] = (frogStates[frogId] + 1) % 5;
    }
    
    for (int i = 0; i < 1000; i++) {
        if (frogExists[i] && frogStates[i] != 0) {
            return 0;
        }
    }
    return 1;
}

int backtrack(char* croak, int index, int* assignment, int maxFrogId, int len) {
    if (index == len) {
        return isValidAssignment(croak, assignment, len) ? maxFrogId + 1 : INT_MAX;
    }
    
    int minFrogs = INT_MAX;
    
    // Try existing frogs
    for (int frogId = 0; frogId <= maxFrogId; frogId++) {
        assignment[index] = frogId;
        int result = backtrack(croak, index + 1, assignment, maxFrogId, len);
        if (result < minFrogs) {
            minFrogs = result;
        }
    }
    
    // Try new frog
    if (maxFrogId < len / 5) {
        assignment[index] = maxFrogId + 1;
        int result = backtrack(croak, index + 1, assignment, maxFrogId + 1, len);
        if (result < minFrogs) {
            minFrogs = result;
        }
    }
    
    return minFrogs;
}

int minNumberOfFrogs(char* croakOfFrogs) {
    int len = strlen(croakOfFrogs);
    if (len == 0 || len % 5 != 0) {
        return -1;
    }
    
    int* assignment = (int*)malloc(len * sizeof(int));
    int result = backtrack(croakOfFrogs, 0, assignment, 0, len);
    free(assignment);
    
    return result == INT_MAX ? -1 : result;
}

Time & Space Complexity

Time Complexity

⏱️

O(5^n)

Each character can be assigned to any of the active frogs, leading to exponential combinations

✓ Linear Growth

Space Complexity

O(n)

Space for recursion stack and storing current assignment

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Assignments) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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