YES or YES? (Codeforces 1703A)
Tags: Implement, strings

There is a string $$$s$$$ of length $$$3$$$, consisting of uppercase and lowercase English letters. Check if it is equal to "YES" (without quotes), where each letter can be in any case. For example, "yES", "Yes", "yes" are all allowable.

Input: The first line of the input contains an integer $$$t$$$ ($$$1 \leq t \leq 10^3$$$) — the number of testcases.The description of each test consists of one line containing one string $$$s$$$ consisting of three characters. Each character of $$$s$$$ is either an uppercase or lowercase English letter.

Output: For each test case, output "YES" (without quotes) if $$$s$$$ satisfies the condition, and "NO" (without quotes) otherwise.You can output "YES" and "NO" in any case (for example, strings "yES", "yes" and "Yes" will be recognized as a positive response).

Note: The first five test cases contain the strings "YES", "yES", "yes", "Yes", "YeS". All of these are equal to "YES", where each character is either uppercase or lowercase.

Sample Inputs:
10
YES
yES
yes
Yes
YeS
Noo
orZ
yEz
Yas
XES

Sample Outputs:
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO

Solution (Python):
t = int(input())
for i in range(t):
    s = input()
    if s.upper() == 'YES':
        print('YES')
    else:
        print('NO')

Solution (Python):
t = int(input())
while t > 0:
    s = input()
    if s.upper() == 'YES':
        print('YES')
    else:
        print('NO')
    t -= 1

Solution (Python):
n = int(input())
while n > 0:
    s = input()
    count = 0
    if s[0] == 'Y' or s[0] == 'y':
        count += 1
    if s[1] == 'E' or s[1] == 'e':
        count += 1
    if s[2] == 'S' or s[2] == 's':
        count += 1
    if count == 3:
        print('YES')
    else:
        print('NO')
    n -= 1