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

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

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minimum complexity to find if a number is present n/2 times or not in a sorted array
Tag : arrays , By : Reiner
Date : November 24 2020, 03:41 PM

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How to calculate the minimum number of person present during two times

Tag : algorithm , By : Zelos
Date : March 29 2020, 07:55 AM
I hope this helps you . Take the start and end times for person number i, and name them s_i and e_i.
Put all of these in a list, times. Then do the following:
# 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)]

complexity in finding the number of times each character is present in a string

Tag : python , By : UnKnownUser
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Yes, O(n) is the best you can do.
It is necessary to visit each character to count them all.
from collections import Counter
input_string = "foobaarfoooobaaaarfo"
counter = Counter(input_string)

Find the minimum of a sorted and shifted array with better than O(n) time complexity

Tag : java , By : Shane
Date : March 29 2020, 07:55 AM
will help you We have an assignment to search for the minimum element of a sorted array that is shifted to the right afterwards. For example: [1, 5, 6, 19, 56, 101] becomes [19, 56, 101, 1, 5, 6]. The method should be implemented using a divide and conquer algorithm and it should have a better asymptotic time complexity than O(n). EDIT: I forgot to add that the elements int the array are unique. , In Java you could use a List because than you can create a Sublist.
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);
        return findMin(rightList);

How can I find a number which occurs an odd number of times in a SORTED array in O(n) time?

Tag : c , By : static AG
Date : March 29 2020, 07:55 AM

given an array of integers in random order you have to find the minimum number of swaps to convert it to cyclic sorted a

Tag : algorithm , By : Kristian Hofslaeter
Date : March 29 2020, 07:55 AM
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