Second most repeated word in a sequence in Python
Last Updated :
10 Apr, 2023
Given a sequence of strings, the task is to find out the second most repeated (or frequent) string in the given sequence. (Considering no two words are the second most repeated, there will be always a single word).
Examples:
Input : {"aaa", "bbb", "ccc", "bbb",
"aaa", "aaa"}
Output : bbb
Input : {"geeks", "for", "geeks", "for",
"geeks", "aaa"}
Output : for
This problem has existing solution please refer Second most repeated word in a sequence link. We can solve this problem quickly in Python using Counter(iterator) method.
Approach is very simple –
- Create a dictionary using Counter(iterator) method which contains words as keys and it’s frequency as value.
- Now get a list of all values in dictionary and sort it in descending order. Choose second element from the sorted list because it will be the second largest.
- Now traverse dictionary again and print key whose value is equal to second largest element.
Implementation
Python3
from collections import Counter
def secondFrequent(input):
dict = Counter(input)
value = sorted(dict.values(), reverse=True)
secondLarge = value[1]
for (key, val) in dict.items():
if val == secondLarge:
print(key)
return
if __name__ == "__main__":
input = ['aaa', 'bbb', 'ccc', 'bbb', 'aaa', 'aaa']
secondFrequent(input)
|
Time complexity: O(nlogn) where n is the length of the input list
Auxiliary space: O(n) where n is the length of the input list
Alternate Implementation :
Python3
from collections import Counter
class Solution:
def secFrequent(self, arr, n):
all_freq = dict(Counter(arr))
store = []
for w in sorted(all_freq, key=all_freq.get):
if w not in store:
store.append(w)
return store[-2]
if __name__ == '__main__':
t = 1
for _ in range(t):
n = 7
arr = ["cat","mat","cat","mat","cat",'ball',"tall"]
ob = Solution()
ans = ob.secFrequent(arr,n)
print(ans)
|
Time complexity: O(nlogn)
Auxiliary space: O(n)
Approach#3: using dictionary
We can use a dictionary to count the frequency of each word in the sequence. Then, we can find the second most repeated word by iterating over the dictionary and keeping track of the maximum and second maximum frequency.
Steps that were to follow the above approach:
- Create an empty dictionary to count the frequency of each word in the sequence.
- Iterate over each word in the sequence and update its frequency in the dictionary.
- Initialize the maximum and second maximum frequency to 0 and -1, respectively.
- Iterate over the items in the dictionary and update the maximum and second maximum frequency if necessary.
- Return the word corresponding to the second maximum frequency.
Python3
def second_most_repeated_word(sequence):
word_count = {}
for word in sequence:
if word in word_count:
word_count[word] += 1
else:
word_count[word] = 1
max_freq = 0
second_max_freq = -1
for word, freq in word_count.items():
if freq > max_freq:
second_max_freq = max_freq
max_freq = freq
elif freq > second_max_freq and freq < max_freq:
second_max_freq = freq
for word, freq in word_count.items():
if freq == second_max_freq:
return word
sequence = ["aaa", "bbb", "ccc", "bbb", "aaa", "aaa"]
print(second_most_repeated_word(sequence))
|
Time complexity: O(n), where n is the number of words in the sequence. This is due to the iteration over each word in the sequence and the items in the dictionary.
Space complexity: O(n), where n is the number of words in the sequence. This is due to the storage of the dictionary.
Please Login to comment...