Second most repeated word in a sequence in Python

07 Jun, 2022
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.
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.
 
Python3
# Python code to print Second most repeated# word in a sequence in Pythonfrom collections import Counter  def secondFrequent(input):     # Convert given list into dictionary    # it's output will be like {'ccc':1,'aaa':3,'bbb':2}    dict = Counter(input)     # Get the list of all values and sort it in ascending order    value = sorted(dict.values(), reverse=True)     # Pick second largest element    secondLarge = value[1]     # Traverse dictionary and print key whose    # value is equal to second large element    for (key, val) in dict.items():        if val == secondLarge:            print(key)            return  # Driver programif __name__ == "__main__":    input = ['aaa', 'bbb', 'ccc', 'bbb', 'aaa', 'aaa']    secondFrequent(input)
Output:  
bbb
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 listAlternate Implementation :  
Python3
# returns the second most repeated wordfrom collections import Counterclass Solution:    def secFrequent(self, arr, n):        all_freq = dict(Counter(arr))        store = []        for w in sorted(all_freq, key=all_freq.get):            # if add key=all_freq.get will sort according to values            # without key=all_freq.get will sort according to keys            if w not in store:                store.append(w)                     return store[-2] # driver code or main functionif __name__ == '__main__':    t = int(input())    for _ in range(t):        n = int(input().strip())        arr = input().strip().split(" ")        ob = Solution()        ans = ob.secFrequent(arr,n)        print(ans)          contributed by Pratyush Pratap Singh
bbb
Time complexity: O(nlogn)
Auxiliary space: O(n)
Midhilesh
pratyushpratapsingh11
hasani
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