Leetcode #1419: Minimum Number of Frogs Croaking
In this guide, we solve Leetcode #1419 Minimum Number of Frogs Croaking in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Counting
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: croakOfFrogs = "croakcroak"
Output: 1
Explanation: One frog yelling "croak" twice.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.