minimum complexity to find if a number is present n/2 times or not in a sorted array

# start_time contains the start time of the event
# end_time contains the start time of the event
sort(times) # When doing this sort, make sure that if a s_i
            # is at the same time as a e_j, put the s_i first.

current_number = number of people who're there at start_time
remove all times that match start_time from times

minimum_number = current_number

for time in times:
    if time is an s_i: current_number += 1

    if time is an e_i:
        current_number -= 1
        if current_number < minimum_number and time < end_time:
            minimum_number = current_number

times = [(s_Bob, 8), (e_Bob, 10), (s_Alice, 10), (e_Alice, 12)]

times = [(s_Bob, 8), (s_Alice, 10), (e_Bob, 10), (e_Alice, 12)]

from collections import Counter
input_string = "foobaarfoooobaaaarfo"
counter = Counter(input_string)
print(counter)

private Integer findMinimum(List<Integer> list) {
    if (list.size() < 2)
        return list.get(0);

    int mid = list.size() / 2;

    // create left and right list
    List<Integer> leftList = list.subList(0, mid);
    List<Integer> rightList = list.subList(mid, list.size());

    if (leftList.get(leftList.size() - 1) <= rightList.get(rightList.size() - 1))
        return findMin(leftList);
    else
        return findMin(rightList);
}