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TruHearing Assessment
const day=[
['8:30', '9:00'],
['9:00', '9:30'],
['9:30', '10:00'],
['10:00', '10:30'],
['10:30', '11:00'],
['11:00', '11:30'],
['11:30', '12:00'],
['1:00', '1:30'],
['1:30', '2:00'],
['2:00', '2:30'],
['2:30', '3:00'],
['3:00', '3:30'],
['3:30', '4:00'],
['4:00', '4:30'],
['4:30', '5:00']
]
const appointments=[
['9:00', '9:30'],
['9:00', '11:30'],
['10:00', '11:00'],
['2:30', '3:00'],
['2:30', '3:30']
]
const availability = [
8.5,
9,
9.5,
10,
10.5,
11,
11.5,
13,
13.5,
14,
14.5,
15,
15.5,
16,
16.5
]
team_availability = () => {
const app = appointments
const avail = availability
//convert appointments
for(let i = 0; i < app.length; i++) {
for(let j = 0; j < app[i].length; j++) {
const split = app[i][j].split(':')
//convert to integers
if(split[1] === '30') {
app[i][j] = parseInt(split[0]) + .5
} else {
app[i][j] = parseInt(split[0])
}
//convert to 24 hr time
if(app[i][j] < 7) {
app[i][j] += 12
}
}
}
//remove same start times
for(let i= 0; i < app.length-1; i++){
if(app[i][0] === app[i+1][0] && app[i][1] <= app[i+1][1]) {
app.splice(i,1)
} else {
app.splice(i+1, 1)
}
}
//check end times and remove overlapping periods
for(let i= 0; i < app.length-1; i++){
if(app[i][1] > app[i+1][1]){
app.splice(i+1, 1)
}
}
//remove unavailable times from availability
for(let i=app.length-1; i >= 0; i--){
let start = avail.indexOf(app[i][0])
let end = avail.indexOf(app[i][1])
day.splice(start, (end - start))
}
return day
}
team_availability()
@SterlingChin
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Author

SterlingChin commented Jul 28, 2017

This can also be run here: repl.it

I have various versions of this assessment saved on repl.it. I chose to go with this option because I'm going to need to explain my thought process to another developer and using more vanilla js allows me to describe it more thoroughly. Prior to going to prod, this code could be cleaned up, comments removed, etc.

I have a (somewhat) cleaned up version using ES6 here: repl.it

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