Last Updated: 11 Dec, 2021
Second Most Repeated Word
Easy
Asked in companies
You are given an array of strings ‘ARR’. You have to find out the second most repeated word in the array ‘ARR’. It is guaranteed every string occurs a unique number of times in the array. If there is only one unique string in the array, return an empty string.
Example:-
N = 5
S = [‘aaa’, ‘bbb’, ‘ccc’, ‘aaa’, ‘bbb’, ‘aaa’]
ANSWER:- The answer should be ‘bbb’ as it is repeated 2 times and is the second most repeated word in the array [after the word ‘aaa’ which is repeated 3 times].
Input Format :
The first line contains a single integer ‘T’ representing the number of test cases. Then each test case follows.
The first line of every test case contains an integer ‘N’ denoting the length of the array.
The next line of every test case contains ‘N’ strings denoting the string in the array.
Output Format :
For each test case, return the second most repeated character in the array. If there is only one unique string in the array, return an empty string.
The output of each test case should be printed in a separate line.
Note :
You are not required to print anything, it has already been taken care of. Just implement the function.
Constraints :
1 <= T <= 10
1 <= N <= 1000
1 <= |ARR[i]| <= 1000
Note:- It is guaranteed that every string has a unique frequency in the array.
Time Limit = 1 sec
Approaches
Iterate through the array and store the frequency of each string in the array. Find out the maximum frequency word which occurs in the array and print the word which has the maximum frequency but has less frequency than the maximum one.
Algorithm:-
- Initialize an empty string named ANS to store the final answer.
- Initialize a dictionary named FREQ to store the frequency of the words in the array.
- Iterate through the array from 0 to N.(Let’s say the iterator is i).
- Increment FREQ[ARR[i]] by 1.
- Initialize a variable named MAX_FREQ with the maximum value of the dictionary FREQ.
- Initialize a variable named SEC_MAX with 0.
- Iterate through the dictionary FREQ.(Let’s say the iterator is i).
- If FREQ[i] is greater than SEC_MAX and but less than MAX_FREQ,
- Update SEC_MAX to FREQ[i] and ANS to i.
- If FREQ[i] is greater than SEC_MAX and but less than MAX_FREQ,
- Return ANS.