Post
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Yonseo

2019-08-05 12:15:24

HackerRank Python Challenge

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Given a range of numbers x and y, how many non repeating numbers are there?

For example: range 219, 220, 221, 222, 223

220, 221, 222, 223 are matching numbers because digit 2 appears twice in each number. 

219 is a non matching number. There is no repeating digit.

Think before you Code

There are 3 problems. First is taking two numbers and creating a range. Lucky for me there's a function for that range(x, y). The next problem is taking each number in the array and converting it into an array because I need to filter out each digit to find a repeating number. In a the array, if it appears more than once it is a match. Finally the last problem is creating an array of non matching numbers and counting that array to return the result.

Code:

def countNumbers(array):

y = range(array[0], array[1] + 1)
if len(y) > 1:
count_item = 0
match_array = []
no_match_array = []
match = False

for item in y:
single_digit_array = [int(d) for d in str(item)]
count_digit = 0
for i in single_digit_array:
single = single_digit_array[count_digit]
z = single_digit_array.count(single)
if z > 1 :
match = True
else:
match = False
count_digit = count_digit + 1
if match == True:
match_array.append(item)
else:
no_match_array.append(item)

count_item = count_item + 1
result = len(no_match_array)
else:
result = None

print 'There are: ' + str(result) + ' non matching numbers'
print no_match_array
print match_array

x = 219
y = 223
array = [x, y]
countNumbers(array)


Result:

There are: 1 non matching numbers

[219]

[220, 221, 222, 223]

Improvements:

I can improve my code by breaking the loop when a match is found so a number like 0012334 will not have to check for a repeating digit if one is already found.


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