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Given a string, you have to find the first n most frequent characters in it. If there are two letters with the same frequency, then the alphabetically earlier value should be picked first:

string=  "aabbccc"
n =2
list = []
#write your code here
char_dict = {}

for char in string:
    if char not in char_dict:
        char_dict[char] = 1
    else:
        char_dict[char] += 1


sorted_dict=sorted(char_dict.items(), key=lambda x: (x[1],x[0]))

sorted_dict = sorted_dict[-2:]
for key, value in sorted_dict:
    list.append(key)
print(list)

My output is ['b', 'c'] but it should actually be c and a.

Adrian Mole
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rinki kumari
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6 Answers6

5

The problem is with the sorting. You need to sort by two fields in different directions (1 ascending and 1 descending). Change the sorted_dict 2 lines to:

sorted_dict = sorted(char_dict.items(), key=lambda x: (-x[1], x[0]))
sorted_dict = sorted_dict[:n]

btw: Avoid using python's keywords (such as list) as variable names. Name it myList or something similar.

Aryerez
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2

Below is one of the perfect solution for your problem(with less code)

string=input()
n=int(input())
import collections
out=[collections.Counter(string).most_common(i+1)[i][0] for i in range(n)]
out.sort()
print(out)
Sunil
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1

I added the print outputs after the statement. The code should be selfdescripting. 

from collections import defaultdict

string=  "aabbccc"
n = 2

result = defaultdict(int)
for char in string:
    result[char] += 1
print(result)  # defaultdict(<class 'int'>, {'b': 2, 'a': 2, 'c': 3})

ordered_result = sorted(result.items(), key=lambda x: (-x[1], x[0]))
print(ordered_result)  # [('c', 3), ('a', 2), ('b', 2)]

ordered_list = [x[0] for x in ordered_result]
print(ordered_list[:n])  # ['c', 'a']
Frank
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0
def char_frequency(string,n):
  letter_freq = dict()
  for letter in string:
    if letter not in letter_freq.keys():
      letter_freq[letter] = 1
    else:
      letter_freq[letter] += 1

  list_of_tuples = sorted(letter_freq.items(), key=lambda x: (-x[1],x[0]))[:n]
  print(list_of_tuples)
  final_list = []
  for tup in list_of_tuples:
    final_list.append(tup[0])
  return(sorted(final_list))

print(char_frequency("aabbccc",2))
Abhishek Sa
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Nov 24 '21 at 02:37
0
string="good"
dictionary = {}
n=2
for char in string:
   if(char in dictionary. keys()):
      dictionary[char]+=1
   else:
      dictionary[char]=1
duplicate=[]
for char in dictionary:
   if(dictionary[char] ==n):
      print(char)
Deva
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  • Remember that Stack Overflow isn't just intended to solve the immediate problem, but also to help future readers find solutions to similar problems, which requires understanding the underlying code. This is especially important for members of our community who are beginners, and not familiar with the syntax. Given that, **can you [edit] your answer to include an explanation of what you're doing** and why you believe it is the best approach? – Jeremy Caney Jan 20 '23 at 00:07
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string="aabbccc"
unique_chars = list(set(string))
count_chars = map(lambda i: string.count(i), unique_chars)

order = sorted(zip(unique_chars, count_chars), key=lambda x: (-x[1], x[0]))
n=2
nth=order[:n]

the variable order is basically mapping each letter to its count

or in 1 line

sorted(zip(list(set(string)), map(lambda i: string.count(i),list(set(string)))), key=lambda x: (-x[1], x[0]))[:2]

The result is in the following format

[('c', 3), ('a', 2)]
rishi
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