LeetCode #1419 — MEDIUM

Minimum Number of Frogs Croaking

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

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 valid "croak" return -1.

Example 1:

Input: croakOfFrogs = "croakcroak"
Output: 1 
Explanation: One frog yelling "croak" twice.

Example 2:

Input: croakOfFrogs = "crcoakroak"
Output: 2 
Explanation: The minimum number of frogs is two. 
The first frog could yell "crcoakroak".
The second frog could yell later "crcoakroak".

Example 3:

Input: croakOfFrogs = "croakcrook"
Output: -1
Explanation: The given string is an invalid combination of "croak" from different frogs.

Constraints:

1 <= croakOfFrogs.length <= 10⁵
croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'.

Step 01

Brute Force Baseline

Problem summary: 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 valid "croak" return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"croakcroak"

Example 2

"crcoakroak"

Example 3

"croakcrook"

Core Insight

What unlocks the optimal approach

keep the frequency of all characters from "croak" using a hashmap.
For each character in the given string, greedily match it to a possible "croak".

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1419: Minimum Number of Frogs Croaking
class Solution {
    public int minNumberOfFrogs(String croakOfFrogs) {
        int n = croakOfFrogs.length();
        if (n % 5 != 0) {
            return -1;
        }
        int[] idx = new int[26];
        String s = "croak";
        for (int i = 0; i < 5; ++i) {
            idx[s.charAt(i) - 'a'] = i;
        }
        int[] cnt = new int[5];
        int ans = 0, x = 0;
        for (int k = 0; k < n; ++k) {
            int i = idx[croakOfFrogs.charAt(k) - 'a'];
            ++cnt[i];
            if (i == 0) {
                ans = Math.max(ans, ++x);
            } else {
                if (--cnt[i - 1] < 0) {
                    return -1;
                }
                if (i == 4) {
                    --x;
                }
            }
        }
        return x > 0 ? -1 : ans;
    }
}

// Accepted solution for LeetCode #1419: Minimum Number of Frogs Croaking
func minNumberOfFrogs(croakOfFrogs string) int {
	n := len(croakOfFrogs)
	if n%5 != 0 {
		return -1
	}
	idx := [26]int{}
	for i, c := range "croak" {
		idx[c-'a'] = i
	}
	cnt := [5]int{}
	ans, x := 0, 0
	for _, c := range croakOfFrogs {
		i := idx[c-'a']
		cnt[i]++
		if i == 0 {
			x++
			ans = max(ans, x)
		} else {
			cnt[i-1]--
			if cnt[i-1] < 0 {
				return -1
			}
			if i == 4 {
				x--
			}
		}
	}
	if x > 0 {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #1419: Minimum Number of Frogs Croaking
class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        if len(croakOfFrogs) % 5 != 0:
            return -1
        idx = {c: i for i, c in enumerate('croak')}
        cnt = [0] * 5
        ans = x = 0
        for i in map(idx.get, croakOfFrogs):
            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 -1 if x else ans

// Accepted solution for LeetCode #1419: Minimum Number of Frogs Croaking
struct Solution;

use std::collections::HashMap;
impl Solution {
    fn min_number_of_frogs(croak_of_frogs: String) -> i32 {
        if croak_of_frogs.len() % 5 != 0 {
            return -1;
        }
        let mut id: HashMap<char, usize> = HashMap::new();
        for &c in &['c', 'r', 'o', 'a', 'k'] {
            id.insert(c, id.len());
        }
        let mut count = vec![0; 5];
        let mut frogs = 0;
        let mut res = 0;
        for c in croak_of_frogs.chars() {
            let i = id[&c];
            count[i] += 1;
            if i == 0 {
                frogs += 1;
                res = res.max(frogs);
            }
            if i > 0 {
                if count[i - 1] < count[i] {
                    return -1;
                }
            }
            if i == 4 {
                frogs -= 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let croak_of_frogs = "croakcroak".to_string();
    let res = 1;
    assert_eq!(Solution::min_number_of_frogs(croak_of_frogs), res);
    let croak_of_frogs = "crcoakroak".to_string();
    let res = 2;
    assert_eq!(Solution::min_number_of_frogs(croak_of_frogs), res);
    let croak_of_frogs = "croakcrook".to_string();
    let res = -1;
    assert_eq!(Solution::min_number_of_frogs(croak_of_frogs), res);
    let croak_of_frogs = "croakcroa".to_string();
    let res = -1;
    assert_eq!(Solution::min_number_of_frogs(croak_of_frogs), res);
}

// Accepted solution for LeetCode #1419: Minimum Number of Frogs Croaking
function minNumberOfFrogs(croakOfFrogs: string): number {
    const n = croakOfFrogs.length;
    if (n % 5 !== 0) {
        return -1;
    }
    const idx = (c: string): number => 'croak'.indexOf(c);
    const cnt: number[] = [0, 0, 0, 0, 0];
    let ans = 0;
    let x = 0;
    for (const c of croakOfFrogs) {
        const i = idx(c);
        ++cnt[i];
        if (i === 0) {
            ans = Math.max(ans, ++x);
        } else {
            if (--cnt[i - 1] < 0) {
                return -1;
            }
            if (i === 4) {
                --x;
            }
        }
    }
    return x > 0 ? -1 : ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.