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First solution in Uncategorized category for Non-unique Elements by Mahoter
#Your optional code here
#You can import some modules or create additional functions
def checkio(data):
#Your code here
#It's main function. Don't remove this function
#It's used for auto-testing and must return a result for check.
result = [0]
cnt = 0
while cnt < len(data):
found = 0
num = 0
for chck in data:
if data[cnt] == chck and cnt != num:
found = 1
if len(str(data[cnt])) == 1 and len(str(chck)) == 1:
if ord(str(data[cnt])) > 96 and ord(str(data[cnt])) < 123 and (ord(str(data[cnt]))-32) == ord(str(chck)) and cnt != num:
found = 1
if ord(str(chck)) > 96 and ord(str(chck)) < 123 and (ord(str(chck))-32) == ord(str(data[cnt])) and cnt != num:
found = 1
num +=1
if found ==1:
result += [data[cnt]]
cnt +=1
cnt = len(result)
result = result[1:cnt+1]
return result
#Some hints
#You can use list.count(element) method for counting.
#Create new list with non-unique elements
#Loop over original list
if __name__ == "__main__":
#These "asserts" using only for self-checking and not necessary for auto-testing
assert isinstance(checkio([1]), list), "The result must be a list"
assert checkio([1, 2, 3, 1, 3]) == [1, 3, 1, 3], "1st example"
assert checkio([1, 2, 3, 4, 5]) == [], "2nd example"
assert checkio([5, 5, 5, 5, 5]) == [5, 5, 5, 5, 5], "3rd example"
assert checkio([10, 9, 10, 10, 9, 8]) == [10, 9, 10, 10, 9], "4th example"
Jan. 17, 2016
