HackerRank Company Logo solution in python | python practice problems for beginners with solutions

A newly opened multinational brand has decided to base their company logo on the three most common characters in the company name. They are now trying out various combinations of company names and logos based on this condition. Given a string , which is the company name in lowercase letters, your task is to find the top three most common characters in the string.

• Print the three most common characters along with their occurrence count.
• Sort in descending order of occurrence count.
• If the occurrence count is the same, sort the characters in alphabetical order.

For example, according to the conditions described above,

 would have it’s logo with the letters .

A single line of input containing the string .

•  has at least  distinct characters

Print the three most common characters along with their occurrence count each on a separate line.
Sort output in descending order of occurrence count.
If the occurrence count is the same, sort the characters in alphabetical order.

Here, b occurs  times. It is printed first.
Both a and c occur  times. So, a is printed in the second line and c in the third line because a comes before c in the alphabet.

Note: The string  has at least  distinct characters.

Problem solution in Python 2 programming.

from sys import stdin

S = stdin.readline()
d = {}
for c in S:
    if c in d:
        d[c] += 1
    else:
        d[c] = 1
        
data = [[d[c],c] for c in d.keys()]
data.sort(key = lambda e: [-e[0],e[1]])

for x in range(3):
    print data[x][1], data[x][0]

#!/bin/python3

import math
import os
import random
import re
import sys
from collections import Counter

class OrderedCounter(Counter):
    pass

if __name__ == '__main__':
    [print(*c) for c in OrderedCounter(sorted(input())).most_common(3)]

# Enter your code here. Read input from STDIN. Print output to STDOUT
from collections import Counter
string = sorted(Counter(raw_input()).items(), key= lambda x: (-x[1],x[0]))[:3]
print "n".join(x[0]+" "+str(x[1]) for x in string)

from collections import Counter, OrderedDict

class OrderedCounter(Counter, OrderedDict):
    pass
[print(*c) for c in OrderedCounter(sorted(input())).most_common(3)]