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 –

1. Create a dictionary using Counter(iterator) method which contains words as keys and it’s frequency as value.
2. 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.
3. Now traverse dictionary again and print key whose value is equal to second largest element.

Implementation

Python3

 `# Python code to print Second most repeated` `# word in a sequence in Python` `from` `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 program` `if` `__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 list

Alternate Implementation :

Python3

 `# returns the second most repeated word` `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 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 function` `if` `__name__ ``=``=` `'__main__'``:` `    ``# no. of test cases` `    ``t ``=` `1` `    ``for` `_ ``in` `range``(t):` `        ``# no of words` `        ``n ``=` `7` `        ``# String of words` `        ``arr ``=` `[``"cat"``,``"mat"``,``"cat"``,``"mat"``,``"cat"``,``'ball'``,``"tall"``]` `        ``ob ``=` `Solution()` `        ``ans ``=` `ob.secFrequent(arr,n)` `        ``print``(ans)`

Output

`mat`

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`   `# Example usage` `sequence ``=` `[``"aaa"``, ``"bbb"``, ``"ccc"``, ``"bbb"``, ``"aaa"``, ``"aaa"``]` `print``(second_most_repeated_word(sequence)) ``# Output: bbb`

Output

`bbb`

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.

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