Q:

My code works well but when I try to search that already included character, it wasn't matching and keeps saying not found

from array import *

alphabets = array('u', [])

lettersRange = int(input("Enter how many letters do you want to enter "))
for i in range(lettersRange):
    addLetter = input("Enter Alphabet ")
    alphabets.append(addLetter)

print("alphabets that you've entered are ", alphabets)

for i in range(len(alphabets)):
    print(alphabets[i])

searchLetter = input("enter the letter to search if its is available or not")
count = 0
for i in alphabets:
    if i == searchLetter:
        print('The Letter "',searchLetter,'" is available and it\'s position is',count )

    else:
        print("Letter Not found")
        break

    count += 1

HERE IS MY OUTPUT:
Enter how many letters do you want to enter 3
Enter Alphabet A
Enter Alphabet B
Enter Alphabet C
alphabets that you've entered are  array('u', 'ABC')
A
B
C
enter the letter to search if its is available or notB
Not found

Also I have attached my output image 

A:

Remove the break, and it will find it. Here is a slightly better code. I have made changes to yours.
from array import *

alphabets = array('u', [])

lettersRange = int(input("Enter how many letters do you want to enter "))
for i in range(lettersRange):
    addLetter = input("Enter Alphabet ")
    alphabets.append(addLetter)

print("alphabets that you've entered are ", alphabets)

for i in range(len(alphabets)):
    print(alphabets[i])

searchLetter = input("enter the letter to search if its is available or not")
count = 0
for i in alphabets:
    if i == searchLetter:
        print('The Letter "',searchLetter,'" is available and it\'s position is',count )
    break
    count += 1

if(count==len(alphabets)):
    print("Not Found")

