Codey and Pebbles

https://www.hackerrank.com/contests/codenection-2024-preliminary-round-closed-category/challenges/cn24-5

Question

Codey and its n friends have gathered in CodeNection Town. Each one of them brings a handful of pebbles they collected by the river. Everyone brings an even or odd number of pebbles, but one friend, by mistake, brings a number that doesn’t match the rest.

The friends line up in a row, each showing the number of pebbles they’ve brought. Help Codey identify the one friend whose pile has a different number type (odd or even) than all the others.

Input Format

The first line contains an integer n, which represents the number of Codey's friends.

The second line contains n space-separated integers, where each integer represents the number of pebbles brought by Codey's friend.

Constraints

$$3 \le n \le 100$$

Output Format

Output a single integer representing the index of the number that differs in evenness, assuming index starts from 1.

Sample Inputs:

Input 0

Input

6
2 4 10 17 8 20

Output

4

Explanation

All numbers are even except 17. The position of 17 is 4, so the output is 4.

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Solution - Binary Check

Given checking even or odd is basically modular 2, the answer should be either 1 and 0.

After that, check if it's only "1" or "0" in an array. Then, locate that odd one out, print its index+1, and done.

Here's my solution:

t = int(input().strip())
n = list(map(int, input().strip().split()))

check_binary = [i % 2 for i in n]

unique_check = 0
if (check_binary.count(1) == 1):
    unique_check = 1
    
for idx, val in enumerate(check_binary):
    if val == unique_check:
        print(idx + 1)

Note that I used the count() feature to count the number of 1s, if the count returns 1, which means the binary "1" is the only cell; otherwise, it's "0".