


You may assume that the sequence is always correct, i.e., every booked room was previously free, and every freed room was previously booked.
In case, 2 rooms have been booked the same number of times, you have to return Lexographically smaller room.
A string 'a' is lexicographically smaller than a string 'b' (of the same length) if in the first position where 'a' and 'b' differ, string 'a' has a letter that appears earlier in the alphabet than the corresponding letter in string 'b'. For example, "abcd" is lexicographically smaller than "acbd" because the first position they differ in is at the second letter, and 'b' comes before 'c'.
n = 6, Arr[] = {"+1A", "+3E", "-1A", "+4F", "+1A", "-3E"}
Now in this example room “1A” was booked 2 times which is the maximum number of times any room was booked. Hence the answer is “1A”.
The first line of input format contains ‘T’ denoting the number of test cases. Then each testcase follows
The first line of each test case contains an integer ‘n’ Denoting the number of times a room is booked or freed.
The second line of the test case contains an array of ‘n’ strings each string denoting which room was booked or freed.
For each test case, Print a string ‘ans’ denoting the room which was booked maximum number of times.
Output for every query will be printed in a separate line.
You are not required to print anything explicitly. It has already been taken care of. Just implement the functions.
1 <= T <= 10
1 <= N <= 10^4
Time Limit: 1 sec
The approach is simple, select a room from the sequence and count the total number of times that room was booked and check this for every room until we reach the end of list.
The steps are as following:
In this approach, we will iterate through the array of strings and if we encounter a room that is booked then we will increase the count of the room by 1 and in the end return the room with a maximum number of bookings.
The steps are as follows: