Company Logo - Hacker Rank Solution

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 .

Input Format

A single line of input containing the string .

Constraints

Output Format

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.

Sample Input 0

aabbbccde

Sample Output 0

b 3
a 2
c 2

Explanation 0

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.

Company Logo - Hacker Rank Solution

This can be solved by using Python counters or counting sort. 

Approach:

STEP 1: Create a list, letters, of size . Initialize each element to .

STEP 2: Traverse through the input string . If a character in the string is 'a', then increment letter by . If 'b' is found, then increment letter by  and so on.

STEP 3: Now run a loop three times. Inside the loop, store the maximum value of letters in the variable max_letter.

STEP 4:. In the scope of the previous loop, run another loop and find the index of the first occurrence of max_letter. Now, print the appropriate letter associated with that index and the max_letter.

S = raw_input()
letters = [0]*26

for letter in S:
    letters[ord(letter)-ord('a')] += 1

for _ in range(3):
    
    max_letter = max(letters)
    
    for index in range(26):
        if max_letter == letters[index]:
            print chr(ord('a')+index), max_letter
            letters[index] = -1
            break

Counter Solution 1

from collections import Counter
from operator import itemgetter

for item in (sorted(sorted(Counter(raw_input()).items()), key = itemgetter(1), reverse = True)[:3]):
    print item[0], item[1]

Counter Solution 2

from collections import Counter

for letter, counts in sorted(Counter(raw_input()).most_common(),key = lambda x:(-x[1],x[0]))[:3]:
    print letter, counts