Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
First solution in Clear category for Pangram by MaciejKrasny
def check_pangram(text):
b=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
k=list("ABCDEFGHIJKLMNOPQRSTUVWXYZ ")
for i in range(0,len(b)):
if not k[i] in text :
if not b[i] in text:
return False
else:
continue
else:
continue
return True
Nov. 18, 2016
