Company Logo in python - HackerRank Solution
Problem :
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 s, 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.
GOOGLE
would have it's logo with the letters
G, O,
E.
Input Format :
A single line of input containing the string S.
Constraints :
- 3 < len(S) <= 10^4
Output Format :
Sort output in descending order of occurrence count.
If the occurrence count is the same, sort the characters in alphabetical
order.
Sample Input :
aabbbccde
Sample Output :
b 3 a 2 c 2
Explanation :
aabbbccde
Here, b occurs 3 times. It is printed first.
Both a and c occur 2 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 S has at least 3 distinct characters.
Solution :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | # Company Logo in python - Hacker Rank Solution # python 3 #!/bin/python3 # Company Logo in python - Hacker Rank Solution START import math import os import random import re import sys import collections if __name__ == '__main__': s = sorted(input().strip()) s_counter = collections.Counter(s).most_common() s_counter = sorted(s_counter, key=lambda x: (x[1] * -1, x[0])) for i in range(0, 3): print(s_counter[i][0], s_counter[i][1]) # Company Logo in python - Hacker Rank Solution END |
Disclaimer :-
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CodeWorld19
Hey, I am Parthe Omkar
What is the need of line 15 here?
Hi :) I actually tried the code without it and it works perfectly fine! I don't think you need it, as you already sort the letters of the string in line 13. However, I am just beginning to teach myself coding, so I am not 100% sure what that line should mean anyways. Hope this helps!
sort when s_counters has 2 or more chars have same count.
Nice Article. thanks for sharing this content
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Without importing those libraries it can be done this way:
if __name__ == '__main__':
x={}
k=[]
p=[]
m=[]
a=[]
b=[]
s = input()
l=list(s)
for i in l:
if i not in k:
k.append(i)
for i in k:
q=l.count(i)
p.append(q)
for i in range(len(k)):
x[k[i]]=p[i]
y=sorted(x)
for i in y:
a.append(i)
b.append(x[i])
for i in range(0,3):
print(a[b.index(max(b))],b[b.index(max(b))])
b[b.index(max(b))]=0
Can you explain; I got lost there?
Thank you for taking the time to write such an informative post. Your blog is not only informative, but it is also very creative.
designing your own logo
I used this, without using the above libraries, but its failing 3/6 test...can some explain why? (I dont have the testcases unlocked so I cant see the reason)
CODE -
def func (s):
s_low = s.lower()
dic = {}
for char in s_low:
if char not in dic:
dic[char] = 0
dic[char]+=1
Keymax_1 = max(dic, key= lambda x: dic[x])
print(f'{Keymax_1} {dic[Keymax_1]}')
del dic[Keymax_1]
Keymax_2 = max(dic, key= lambda x: dic[x])
print(f'{Keymax_2} {dic[Keymax_2]}')
del dic[Keymax_2]
Keymax_3 = max(dic, key= lambda x: dic[x])
print(f'{Keymax_3} {dic[Keymax_3]}')