Hi, your solution works and I cant really find any difference from mine exept that in yours is used removeIf lambda, yet still your solution passes and mine not. Probably some Java related thing which is beyond my knowledge scope. Because my for cycle and your removeIf should do exactly the same, isnt it?
I dont know, its so hard to do any testing if tests in IDE are different from “submit”. I understand protection from hard-coders, but if there is no hardcode and it passes in IDE, then it shuld be fine… Or at least if submit tests give some feedback.
I have absolutely no idea how to improve in this puzzle with this kind of behavior and testing 
Anyway, thank you for checking it out.
Cheers,
Petr
2 Likes
Get to level 29 and then you’ll see the validators for the training puzzles 
1 Like
Aha, so one have to go blind and is alowed to see once he gets to where is seeing not relevant anymore, how esoteric 
But thank you, good to know.
cheers
anyone had already solved the big heist problem? I can’t seem to understand the said logic:
“Once he finishes one vault, he moves on to the next vault of greater index (robbers work the vaults in increasing order) which nobody has worked on yet”
Hi.
I trying to solve the puzzle but can’t even calculate the example:
Input
1
1
3 1
Here is my solution:
// the password: 1 digit + 2 chars (3 - 1)
// count of possible digits = 10 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
// count of possible chars = 5 ['A', 'E', 'I', 'O', 'U']
time_for_digits = 10^1 = 10
time_for_chars = 5^2 + 5^1 = 30
time_total = 10 * 30 = 300
But the right answer is 250.
Need some hints…
Nope, error here.
Your are right. Wrong formula. I got it, thank you.
Hi? My thoughts about this:
1 for number of robbers
1 for number of vaults
3 kinds of combination with 1 of it has a combination from 0-9 and the other are from A,E,I,O,U
so getting the total combination will be 10 * 5 * 5 = 250
I solved this puzzle and get green for all testcases, but when I submit it I only get a 50% score? How is this possible?
2 Likes
I think this puzzle has validator issues. The tests are weaker than the validators, so that a wrong code will pass them all and get stuck on the validation, leaving the coder with no clue about what is wrong with their algorithm.
To offset the punishing effect of these validators, I’ll give a few clues about what is expected.
An example that might help understand the problem:
Let’s say we have just 3 robbers (Andy, Bob and Carl) and 10 vaults numbered 1 to 10
All vaults have very simple codes that take one minute to break, except vault 4 that takes 2 minutes and vault 6 that takes 29 minutes.
At first Andy, Bob and Carl work on locks 1 to 3 and take one minute each to crack them.
Then Andy gets to work on lock 4, Bob on lock 5 and Carl on lock 6. Andy will get stuck for 2 minutes, Carl for nearly half an hour. In the mean time, Bob will have time to crack two locks. Then Andy and Bob will get asigned vaults in quick succession until none remains to be cracked. Both will twiddle their thumbs until Carl finally breaks vault 6’s lock.
Each time a lock gets cracked, the thief that is done with it gets reallocated to a new vault (if any remains to be cracked). The rules for reallocation are clearly stated in the problem definition.
Each thief will spend a total time equal to the sum of the time needed to crack each vault that was assigned to him. The final time is simply the max of the times spent by each thief when the last lock is cracked.
Here is some pseudo-code that might help get an idea of the solution.
There are other ways to look at the problem, this is just the simplest to explain I could think of.
Each thief is assigned a cumulated time counter, initially zero.
Each thief begins with no vault assigned to him.
- while not all vaults are cracked
- if there are still vaults available (not cracked yet and with no thief working on them)
- look for an available thief (his cumulated time will be the shortest)
- reassign the thief to a new vault and increase his cumulated time by that of the new vault
- count one more cracked vault
- if there are still vaults available (not cracked yet and with no thief working on them)
- final time is the longest cumulated time.
In the example:
All thieves initially available Andy -> vault 1, time 0:01, Bob & Carl available Bob -> vault 2, time 0:01, Carl available Carl -> vault 3, time 0:01, Andy, Bob & Carl available Andy -> vault 4, time 0:03, Bob & Carl available Bob -> vault 5, time 0:02, Carl available Carl -> vault 6, time 0:30, Bob available Bob -> vault 7, time 0:03, Andy & Bob available Andy -> vault 8 time 0:04, Bob available Bob -> vault 9, time 0:04, Andy & Bob available Andy -> vault 10, time 0:05, no more vaults to crack final time : max (0:05, 0:04, 0:30) = 30 minutes.
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feel free to suggest modifications in the IDE tests.
1 Like
For one thing, out of the 4 tests only the last one exercises the code in any meaningful way.
As it turns out, it’s the first robber who spends the most time and holds the final result.
There are no cases of more than one thief being available at the same time except during initial vault attribution.
All these are potential pitfalls for awkward code. At the very least you should have more than one “big heist” test, and possibly carefully thought out values to make sure vault attribution works as intended.
If anything else fails, swapping the validatoirs with the tests could do the trick.
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ok this is my code well the evaluation. befor that i looped through all vaults to calculate the time needed for each vailt. the values are stored in a list called vaults
so from there I do
//loop through all robbers the first round to give them all a vault
for (var j = 0; j < R; ++j) {
int vaultTime = vaults.removeAt(0);
robbers[j] = robbers[j] + vaultTime;
stderr.writeln('robbers[$j] ${robbers[j]}');
}
robbers.sort();//sort for follow up vault
//while robber with lowest time is lower than robber with secondlowest time
while(robbers[0]<robbers[1] && vaults.isNotEmpty){
int vaultTime = vaults.removeAt(0);
//add vault time to robber
robbers[0] = robbers[0] + vaultTime;
stderr.writeln('1robbers ${robbers} adding $vaultTime');
//sort robbers to see who is free next
if(robbers[0]>robbers[1]){
robbers.sort();
}
}
total = robbers.reduce(max);//(takes biggest value)
my output is
vaults [125000, 156250, 50000, 6250000, 125, 5000, 781250, 5000, 500000, 5000000, 62500, 25000, 50000, 78125, 78125, 6250000, 15625, 500000, 500, 2500]
loop through robers
//first round
robers[0] 125000
robers[1] 156250
robers[2] 50000
robers[3] 6250000
robers[4] 125
//follow up
1robers [5125, 50000, 125000, 156250, 6250000] added 5000
1robers [786375, 50000, 125000, 156250, 6250000] added 781250
1robers [55000, 125000, 156250, 786375, 6250000] added 5000
1robers [555000, 125000, 156250, 786375, 6250000] added 500000
1robers [5125000, 156250, 555000, 786375, 6250000] added 5000000
1robers [218750, 555000, 786375, 5125000, 6250000] added 62500
1robers [243750, 555000, 786375, 5125000, 6250000] added 25000
1robers [293750, 555000, 786375, 5125000, 6250000] added 50000
1robers [371875, 555000, 786375, 5125000, 6250000] added 78125
1robers [450000, 555000, 786375, 5125000, 6250000] added 78125
1robers [6700000, 555000, 786375, 5125000, 6250000] added 6250000
1robers [570625, 786375, 5125000, 6250000, 6700000] added 15625
1robers [1070625, 786375, 5125000, 6250000, 6700000] added 500000
1robers [786875, 1070625, 5125000, 6250000, 6700000] added 500
1robers [789375, 1070625, 5125000, 6250000, 6700000] added 2500
6700000
any guess where i am going wrong?
…? Dafuq?
My initial robber array looks like this:
125:10000:125000:156250:6250000:
Note the “10000” instead of your “50000”…
You’re talking about test case 04 “Big Heist” correct?
And I get different vault values in some places, too. I notice a factor 5 difference for two vaults that I have as “1000” and “100000”…
Can you double check your input routine?
1 Like
hmm thank you!
I fixed a bug where I worked with C as the length of letters… should have been C-N …
well at least I thought that was the problem… now im passing 100% of the tests in the browser. but once submitted i only reach 50%.
I just solved it today (or yesterday since it just passed midnight), and my guess is that since you pass the test cases your algorithm is probably correct, cause the last test would probably make you fail otherwise. So, a hint is to think about avoiding timeout and optimize how you step through the “work” each robber does on the vault.
1 Like
found the bug
removed the whole “robbers[0]>robbers[1]” thing since i sort the first is always the lowest ^^ thanks. 100% <3
cookies for everyone:cookie: 1UP
2 Likes
thanks and congrats! that was just my thought: maybe optimise speed. then i found a unneeded <
maybe it would have worked with a <= but without even better 
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I got stuck at last test case. Got “6473750” instead “6515625”
Does any one have any idea?
My code
const R = parseInt(readline());
const V = parseInt(readline());
let combinations = []
for (let i = 0; i < V; i++) {
var inputs = readline().split(’ ');
const C = parseInt(inputs[0]);
const N = parseInt(inputs[1]);
let combine = calcCombination(C, N)
combinations.push(combine)
}function calcCombination(c, n) {
let d = c - n
return Math.pow(10, n) * Math.pow(5, d)
}// Write an action using console.log()
// To debug: console.error(‘Debug messages…’);combinations = combinations.sort((a, b) => a - b)
let time = 0
for (let i = 0; i < combinations.length; i++) {
let combine = combinations[i]
time += combine
for (let j = 0; j < R; j++) {
if (combinations[i+j]) {
combinations[i+j] -= combine
}
}
}console.log(time);
Thank you, ends up I was being a helluva lot more efficient’n what the prompt was asking for, so my robbers were much faster’n the validator’s dumb robbers LuL