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 : {"Lazyroar", "for", "Lazyroar", "for",
"Lazyroar", "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
# Python code to print Second most repeated# word in a sequence in Pythonfrom collections import Counterdef 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) |
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 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__': # 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) |
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 usagesequence = ["aaa", "bbb", "ccc", "bbb", "aaa", "aaa"]print(second_most_repeated_word(sequence)) # Output: bbb |
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.
