Discusson about Zoos (Problem from HackerEarth)

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Difficulty: Easy

For this problem:

Problem
You are required to enter a word that consists of x and y that denote the number of Zs and Os respectively. The input word is considered similar to word zoo if 
2*x = y.

Determine if the entered word is similar to word zoo.

For example, words such as zzoooo and zzzoooooo are similar to word zoo but not the words such as zzooo and zzzooooo.

Input format

First line: A word that starts with several Zs and continues by several Os.
Note: The maximum length of this word must be 20.

Output format
Print Yes if the input word can be considered as the string zoo otherwise, print No.

Sample Input
zzzoooooo

Sample Output
Yes

I wrote the following code to solve the problem:

name=input()

var_z=0
var_o=0

var_z = name.count("z")
var_o = name.count("o")

if var_o == 2 * var_z:
    print("Yes")
else:
    print("No")

Can you critique my code and suggest any improvements or optimizations? Also, please provide a brief explanation of how the code works and any potential edge cases I should consider.

Your code is correct, concise, and easy to understand for the given problem. But let's go through a detailed critique, suggestions for improvements, and a review of possible edge cases.

---

✅ How Your Code Works

python
name = input()

var_z = name.count("z")
var_o = name.count("o")

if var_o == 2 * var_z:
    print("Yes")
else:
    print("No")

Explanation:

• name.count("z") counts the number of 'z' characters → stored in var_z.

• name.count("o") counts the number of 'o' characters → stored in var_o.

• It then checks if var_o == 2 * var_z as per the condition in the problem.

---

✅ Pros

• ✅ Correct logic: Accurately implements the 2 * z = o check.

• ✅ Simple and readable: Easy to understand even for beginners.

• ✅ Short and efficient: count() is a fast operation for small strings (within constraints).

---

🔍 Suggestions for Improvement

1. Check for invalid characters (input validation)

Right now, the code assumes that input will only contain z and o. It's good to validate that assumption:

python
if set(name) - {'z', 'o'}:
    print("Invalid input: Only 'z' and 'o' allowed.")
else:
    ...

2. Use better variable names

Avoid var_z and var_o — use names that reflect meaning:

python
num_z = name.count('z')
num_o = name.count('o')

3. Efficiency in very long strings

Although your approach is already efficient for normal input sizes, if the input is extremely large (e.g. 10^6 characters), you could avoid traversing the string twice:

python
num_z = num_o = 0
for ch in name:
    if ch == 'z':
        num_z += 1
    elif ch == 'o':
        num_o += 1
    else:
        print("Invalid input")
        exit()

This version processes the string once and handles invalid characters too.

---

⚠️ Edge Cases to Consider

Input	Expected Output	Notes
zzoooo	Yes	Standard valid case
zzooo	No	Fails 2z = o
zoo	Yes	1z, 2o
zozo	No	Invalid character order
zzzooooo	No	3z vs 5o → 5 ≠ 6
zzzoooooo	Yes	3z vs 6o
zzzzzz	No	6z vs 0o
oooooo	No	0z vs 6o
Empty input	No	or handle as invalid
zOo	No	Case-sensitive: 'O' ≠ 'o'

---

✅ Final Polished Version

python
name = input().strip()

if set(name) - {'z', 'o'}:
    print("No")
else:
    num_z = name.count('z')
    num_o = name.count('o')

    if num_o == 2 * num_z:
        print("Yes")
    else:
        print("No")

---

🧠 Summary

• Your code works correctly and efficiently.

• You can improve variable naming, optionally add input validation, and handle edge cases like mixed characters or casing.

• For maximum robustness, scan the string once instead of using .count() twice — though for short strings it's not a concern.

Let me know if you want a version that reads from a file or handles multiple test cases!

Tags: Technology,Algorithms,