1. Rotate upper letters and lower letters in a string

Difficulity: ✩

Give a string s, rotate all lower letters (a-z) to upper letters (A-Z), and process the same way for upper letters. Input: ‘uuuCCKsf(.’ Output: ‘UUUcckSF(.’

Use Python string library could provide a straightforward solution

import string
def rotate_letter(s):
    l = list(s)
    for index, letter in enumerate(l):
        if letter.islower():
            l[index] = letter.upper()
        elif letter.isupper():
            l[index] = letter.lower()
    return ''.join(l)
if __name__ == '__main__':
    s = input()
    rotate_letter(s)

2. Statistic letter show times in a string

Difficulity: ✩✩

Give a string s, make a statistic of letters and order by frequency, if several letters have the same frequency, order by natural order (a-z) Input: hufosniuosfbyvgugbyuceca Output: u:4; b:2; c:2; f:2; g:2; o:2; s:2; y:2; a:1; e:1; h:1; i:1; n:1; v:1;

Firstly, for python, this is a hash problem, make a dict and a counter will help to make this statistic.

>>>> input: 'hufosniuosfbyvgugbyuceca'
s = input()
c = Counter(s)
print(c)
>>>> ouput: Counter({'h': 1, 'u': 4, 'f': 2, 'o': 2, 's': 2, 'n': 1, 'i': 1, 'b': 2, 'y': 2, 'v': 1, 'g': 2, 'c': 2, 'e': 1, 'a': 1})

Secondly, let’s do the sorting part. Sorting a list or dictionary with only one key could be easy.

# sort the dict by value in desc
print(sorted(c.items(), key=lambda x: x[1], reverse=True))
>>>> output: [('u', 4), ('f', 2), ('o', 2), ('s', 2), ('b', 2), ('y', 2), ('g', 2), ('c', 2), ('h', 1), ('n', 1), ('i', 1), ('v', 1), ('e', 1), ('a', 1)]

We may notice that the the order of keys with same value are still inn their string’s first show order ('f', 2), ('o', 2), ('s', 2), ('b', 2), ('y', 2), ('g', 2), ('c', 2). Then here comes the problem with ordering a list / dictionary with two keys. Plus, the second key is not numeric. To deal with this problem. We can use ord() to convert a letter into it’s ascii number. Then, we use two keys in the sort part.

sorted_c = sorted(c.items(), key=lambda x: (x[1],-ord(x[0])), reverse=True)
print(c)
>>>> output: [('u', 4), ('b', 2), ('c', 2), ('f', 2), ('g', 2), ('o', 2), ('s', 2), ('y', 2), ('a', 1), ('e', 1), ('h', 1), ('i', 1), ('n', 1), ('v', 1)]

Finally, decorate the output part as required.

output = []
for item in sorted_c:
    output.append(str(item[0]) + ':' + str(item[1]) + ';')
print(output)
>>>> output: u:4; b:2; c:2; f:2; g:2; o:2; s:2; y:2; a:1; e:1; h:1; i:1; n:1; v:1;

Now, we are done~

3 Decimal to letter

Difficulity: ✩✩ Same as Leetcode 168 Excel Sheet Column Title.

Given a list of positive integers, return a list of corresponding column titles as appear in an Excel sheet. Translate rule: 1 -> A, 2 -> B, 3 -> C … 26 -> Z, 27 -> AA, 28 -> AB …. Input: [1, 28, 701] Output: [“A”, “AB”, “ZY”]

Let’s see the relationship between the Excel sheet column title and the number:

A   1     AA    26+ 1     BA  2×26+ 1     ...     ZA  26×26+ 1     AAA  1×26²+1×26+ 1
B   2     AB    26+ 2     BB  2×26+ 2     ...     ZB  26×26+ 2     AAB  1×26²+1×26+ 2
.   .     ..    .....     ..  .......     ...     ..  ........     ...  .............   
.   .     ..    .....     ..  .......     ...     ..  ........     ...  .............
.   .     ..    .....     ..  .......     ...     ..  ........     ...  .............
Z  26     AZ    26+26     BZ  2×26+26     ...     ZZ  26×26+26     AAZ  1×26²+1×26+26

Now we can see that ABCD＝A×26³＋B×26²＋C×26¹＋D＝1×26³＋2×26²＋3×26¹＋4 But how to get the column title from the number? We can’t simply use the n%26 method because: ZZZZ＝Z×26³＋Z×26²＋Z×26¹＋Z＝26×26³＋26×26²＋26×26¹＋26 We can use (n-1)%26 instead, then we get a number range from 0 to 25.

Code:

def convertToTitle(n: int) -> str:
    import string
    letters = list(string.ascii_uppercase)
    result = []
    if n == 0:
        return ""
    while n > 0:
        result.append(letters[(n-1) % 26])
        n = (n - 1) // 26
    return ''.join(result[::-1])

if __name__ == "__main__":
    numbers = input()
    return [convertToTitle(number) for number in numbers]

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