The hackerrank challenge is in the following url: https://www.hackerrank.com/challenges/python-time-delta/problem
I got testcase 0 correct, but the website is saying that I have wrong answers for testcase 1 and 2, but in my pycharm, I copied the website expected output and compared with my output and they were exactly the same.
Please have a look at my code.
#!/bin/pyth
# Complete the time_delta function below.
from datetime import datetime
def time_delta(tmp1, tmp2):
dicto = {'Jan':1, 'Feb':2, 'Mar':3,
'Apr':4, 'May':5, 'Jun':6,
'Jul':7, 'Aug':8, 'Sep':9,
'Oct':10, 'Nov':11, 'Dec':12}
# extracting t1 from first timestamp without -xxxx
t1 = datetime(int(tmp1[2]), dicto[tmp1[1]], int(tmp1[0]), int(tmp1[3][:2]),int(tmp1[3][3:5]), int(tmp1[3][6:]))
# extracting t1 from second timestamp without -xxxx
t2 = datetime(int(tmp2[2]), dicto[tmp2[1]], int(tmp2[0]), int(tmp2[3][:2]), int(tmp2[3][3:5]), int(tmp2[3][6:]))
# converting -xxxx of timestamp 1
t1_utc = int(tmp1[4][:3])*3600 + int(tmp1[4][3:])*60
# converting -xxxx of timestamp 2
t2_utc = int(tmp2[4][:3])*3600 + int(tmp2[4][3:])*60
# absolute difference
return abs(int((t1-t2).total_seconds()-(t1_utc-t2_utc)))
if __name__ == '__main__':
# fptr = open(os.environ['OUTPUT_PATH'], 'w')
t = int(input())
for t_itr in range(t):
tmp1 = list(input().split(' '))[1:]
tmp2 = list(input().split(' '))[1:]
delta = time_delta(tmp1, tmp2)
print(delta)
t1_utc = int(tmp1[4][:3])*3600 + int(tmp1[4][3:])*60
For a time zone like +0715, you correctly add “7 hours of seconds” and “15 minutes of seconds”
For a timezone like -0715, you are adding “-7 hours of seconds” and “+15 minutes of seconds”, resulting in -6h45m, instead of -7h15m.
You need to either use the same “sign” for both parts, or apply the sign afterwards.
Related
When running the Function with no names on the input lists, it gives everyone the approtriate time based on the varible listed. If we have the input varibale have a name it gives everyone time off instead of just that indivudual.
There are some lists associated with this as well and that part works fine.
this is the required resources:
from openpyxl import Workbook
from datetime import timedelta, datetime
import random
def add_agents_w1():
num = 4
c = ["B","C","D","E","F"]
tow1 = input("IF anyone taking time off enter 1st
person now: \n")
tow12 = input("If someone else is taking time off enter
2nd person now: \n")
for x in tl:
ws1[f"A{num}"] = x
ws1[f"I{num}"] = x
for f in c:
if x in tow1 or tow12:
ws1[f'{f}{num}'] = "OFF"
else:
ws1[f"{f}{num}"] = s2t
num += 1
wb.save(dest_filename)
Hi I would like to subtract time from a CSV array using Ruby
time[0] is 12:12:00AM
time[1] is 12:12:01AM
Here is my code
time_converted = DateTime.parse(time)
difference = time_converted[1].to_i - time_converted[0].to_i
p difference
However, I got 0
p time[0].to_i gives me 12
is there a way to fix this?
You can use Time#strptime to define the format of the parsed string.
In your case the string is %I:%M:%S%p.
%I = 12 hour time
%M = minutes
%S = seconds
%p = AM/PM indicator
So to parse your example:
require 'time'
time = %w(12:12:00AM 12:12:01AM)
parsed_time = time.map { |t| Time.strptime(t, '%I:%M:%S%p').to_i }
parsed_time.last - parsed_time.first
=> 1
Use the Ruby DateTime class and parse your dates into objects of that class.
I am studying for this great Coursera course https://www.coursera.org/learn/algorithmic-toolbox . On the fourth week, we have an assignment related to binary trees.
I think I did a good job. I created a binary search code that solves this problem using recursion in Python3. That's my code:
#python3
data_in_sequence = list(map(int,(input().split())))
data_in_keys = list(map(int,(input()).split()))
original_array = data_in_sequence[1:]
data_in_sequence = data_in_sequence[1:]
data_in_keys = data_in_keys[1:]
def binary_search(data_in_sequence,target):
answer = 0
sub_array = data_in_sequence
#print("sub_array",sub_array)
if not sub_array:
# print("sub_array",sub_array)
answer = -1
return answer
#print("target",target)
mid_point_index = (len(sub_array)//2)
#print("mid_point", sub_array[mid_point_index])
beg_point_index = 0
#print("beg_point_index",beg_point_index)
end_point_index = len(sub_array)-1
#print("end_point_index",end_point_index)
if sub_array[mid_point_index]==target:
#print ("final midpoint, ", sub_array[mid_point_index])
#print ("original_array",original_array)
#print("sub_array[mid_point_index]",sub_array[mid_point_index])
#print ("answer",answer)
answer = original_array.index(sub_array[mid_point_index])
return answer
elif target>sub_array[mid_point_index]:
#print("target num higher than current midpoint")
beg_point_index = mid_point_index+1
sub_array=sub_array[beg_point_index:]
end_point_index = len(sub_array)-1
#print("sub_array",sub_array)
return binary_search(sub_array,target)
elif target<sub_array[mid_point_index]:
#print("target num smaller than current midpoint")
sub_array = sub_array[:mid_point_index]
return binary_search(sub_array,target)
else:
return None
def bin_search_over_seq(data_in_sequence,data_in_keys):
final_output = ""
for key in data_in_keys:
final_output = final_output + " " + str(binary_search(data_in_sequence,key))
return final_output
print (bin_search_over_seq(data_in_sequence,data_in_keys))
I usually get the correct output. For instance, if I input:
5 1 5 8 12 13
5 8 1 23 1 11
I get the correct indexes of the sequences or (-1) if the term is not in sequence (first line):
2 0 -1 0 -1
However, my code does not pass on the expected running time.
Failed case #4/22: time limit exceeded (Time used: 13.47/10.00, memory used: 36696064/536870912.)
I think this happens not due to the implementation of my binary search (I think it is right). Actually, I think this happens due to some inneficieny in a peripheral part of the code. Like the way I am managing to output the final answer. However, the way I am presenting the final answer does not seem to be really "heavy"... I am lost.
Am I not seeing something? Is there another inefficiency I am not seeing? How can I solve this? Just trying to present the final result in a faster way?
TL;DR: I need to get the difference between HH:MM:SS.ms and HH:MM:SS.ms as HH:MM:SS:ms
What I need:
Here's a tricky one. I'm trying to calculate the difference between two timestamps such as the following:
In: 00:00:10.520
Out: 00:00:23.720
Should deliver:
Diff: 00:00:13.200
I thought I'd parse the times into actual Time objects and use the difference there. This works great in the previous case, and returns 00:0:13.200.
What doesn't work:
However, for some, this doesn't work right, as Ruby uses usec instead of msec:
In: 00:2:22.760
Out: 00:2:31.520
Diff: 00:0:8.999760
Obviously, the difference should be 00:00:8:760 and not 00:00:8.999760. I'm really tempted to just tdiff.usec.to_s.gsub('999','') ……
My code so far:
Here's my code so far (these are parsed from the input strings like "0:00:10:520").
tin_first, tin_second = ins.split(".")
tin_hours, tin_minutes, tin_seconds = tin_first.split(":")
tin_usec = tin_second * 1000
tin = Time.gm(0, 1, 1, tin_hours, tin_minutes, tin_seconds, tin_usec)
The same happens for tout. Then:
tdiff = Time.at(tout-tin)
For the output, I use:
"00:#{tdiff.min}:#{tdiff.sec}.#{tdiff.usec}"
Is there any faster way to do this? Remember, I just want to have the difference between two times. What am I missing?
I'm using Ruby 1.9.3p6 at the moment.
Using Time:
require 'time' # Needed for Time.parse
def time_diff(time1_str, time2_str)
t = Time.at( Time.parse(time2_str) - Time.parse(time1_str) )
(t - t.gmt_offset).strftime("%H:%M:%S.%L")
end
out_time = "00:00:24.240"
in_time = "00:00:14.520"
p time_diff(in_time, out_time)
#=> "00:00:09.720"
Here's a solution that doesn't rely on Time:
def slhck_diff( t1, t2 )
ms_to_time( time_as_ms(t2) - time_as_ms(t1) )
end
# Converts "00:2:22.760" to 142760
def time_as_ms( time_str )
re = /(\d+):(\d+):(\d+)(?:\.(\d+))?/
parts = time_str.match(re).to_a.map(&:to_i)
parts[4]+(parts[3]+(parts[2]+parts[1]*60)*60)*1000
end
# Converts 142760 to "00:02:22.760"
def ms_to_time(ms)
m = ms.floor / 60000
"%02i:%02i:%06.3f" % [ m/60, m%60, ms/1000.0 % 60 ]
end
t1 = "00:00:10.520"
t2 = "01:00:23.720"
p slhck_diff(t1,t2)
#=> "01:00:13.200"
t1 = "00:2:22.760"
t2 = "00:2:31.520"
p slhck_diff(t1,t2)
#=> "00:00:08.760"
I figured the following could work:
out_time = "00:00:24.240"
in_time = "00:00:14.520"
diff = Time.parse(out_time) - Time.parse(in_time)
Time.at(diff).strftime("%H:%M:%S.%L")
# => "01:00:09.720"
It does print 01 for the hour, which I don't really understand.
In the meantime, I used:
Time.at(diff).strftime("00:%M:%S.%L")
# => "00:00:09.720"
Any answer that does this better will get an upvote or the accept, of course.
in_time = "00:02:22.760"
out_time = "00:02:31.520"
diff = (Time.parse(out_time) - Time.parse(in_time))*1000
puts diff
OUTPUT:
8760.0 millliseconds
Time.parse(out_time) - Time.parse(in_time) gives the result in seconds so multiplied by 1000 to convert into milliseconds.
I have one date, let's say '2010-12-20' of a flight departure, and two times, for instance, '23:30' and '02:15'.
The problem: I need to get datetimes (yyyy-MM-dd HH:mm:ss, for example, 2010-12-17 14:38:32) of both of these dates, but I don't know the day of the second time (it can be the same day as departure, or the next one).
I am looking for the best solution in Ruby on Rails. In PHP would just use string splitting multiple times, but I believe, that Rails as usually, has a much more elegant way.
So, here is my pseudo code, which I want to turn into Ruby:
depart_time = '23:30'
arrive_time = '02:15'
depart_date = '2010-12-20'
arrive_date = (arrive.hour < depart.hour and arrive.hour < 5) ? depart_date + 1 : depart_date
# Final results
depart = depart_date + ' ' + depart_time
arrive = arrive_date + ' ' + arrive_time
I want to find the best way to implement this in Ruby on Rails, instead of just playing with strings.
This is just pure Ruby, nothing to do with Rails:
require 'date'
depart_time = DateTime.strptime '23:30', '%H:%M'
arrive_time = DateTime.strptime '02:15', '%H:%M'
arrive_date = depart_date = Date.parse( '2010-12-20' )
arrive_date += 1 if arrive_time.hour < depart_time.hour and arrive_time.hour < 5
puts "#{depart_date} #{depart_time.strftime '%H:%M'}",
"#{arrive_date} #{arrive_time.strftime '%H:%M'}"
#=> 2010-12-20 23:30
#=> 2010-12-21 02:15