Problem 62. Elapsed Time

Created by Cody Team

Solve Later

Given two date strings d1 and d2 of the form yyyy/mm/dd HH:MM:SS (assume hours HH is in 24 hour mode), determine how much time, in decimal hours, separates them. Assume d2 is always later than d1.

Example:

 Input d1 = '2010/12/14 12:00:00'
 Input d2 = '2010/12/14 13:06:36'
 Output elapsed is 1.11

Solve

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Last Solution submitted on Dec 06, 2023

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Bobby Cheng on 27 Jan 2012

The testsuite input does not following the input format string.

Chew Baca on 13 Mar 2013

Having a hard time with the test suite regarding the number of hours in one year. I think it should be 8 765.81277h/y

Matthew Kishe on 15 Aug 2013

The test suite doesnt follow. it should be double digit days and month, not single digits according to his descrption

Hassan Javaid on 5 Sep 2013

having problems with the test suite especially test no 3

Gareth Lee on 10 Nov 2014

having problems with test no 3 of the test suite,why?

Riyasat on 5 Feb 2015

It's a fun problem

huashou on 11 Apr 2017

the test suits are all passed , but the mathworks net tells its wrong.
and the size is '-1'.
i use the datenum function.

Chen Qian on 11 Jun 2019

I use format bank to get 2 decimal output, anyone has a better idea?

Sanzhar Askaruly on 6 Jul 2019

using existing functions makes this trick easy

Laza Álmos on 27 Mar 2020

with an existing funcion it is solveable

Bugrahan Atar on 9 Feb 2021

function elapsed= elapsed_time(d1,d2)
d1='2010/12/14 12:00:00'
d2='2010/12/15 13:06:36'
a=(str2num(d2(1:4))-str2num(d1(1:4)))*8760;
b=(str2num(d2(6:7))-str2num(d1(6:7)))*720;
c=(str2num(d2(9:10))-str2num(d1(9:10)))*24;
e=str2num(d2(12:13))-str2num(d1(12:13));
f=(str2num(d2(15:16))-str2num(d1(15:16)))/60;
g=(str2num(d2(18:19))-str2num(d1(18:19)))/3600;
elapsed=a+b+c+e+f+g
end

it told me that solution is wrong?

Tân Trần on 7 Jul 2021

Nice question

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Problem 62. Elapsed Time

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