Timeline for "Open and closed doors" riddle
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2014 at 9:42 | vote | accept | Anirudh | ||
| Aug 28, 2014 at 14:15 | history | edited | Pimgd | CC BY-SA 3.0 |
MATHJAAAX
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| Aug 28, 2014 at 14:11 | comment | added | Anirudh | Yeap! I think you are right. I probably made a mistake while evaluating Iterations :-] | |
| Aug 28, 2014 at 14:07 | comment | added | Pimgd | @Anirudh \$O(n!)\$? I disagree... you had \$O(n^2)\$ with extra reductions. N = 10 for instance, you'd have 10 + (10/2 = 5) + (10/3 = 3) + (10/4 = 2) + (10/5 = 2) + (10/6 = 1) + 1 + 1 + 1 + 1 = 27, not 3.8 million. | |
| Aug 28, 2014 at 14:04 | comment | added | Anirudh | @Pimgd Well you have simplified this pure on the basis of optimizing the Algorithm which works better Space and Memory complexity wise with \$O(n)\$ Time complexity and \$O(1)\$ Space Complexity. As compared to my Time Complexity \$O(n!)\$ and Space \$O(n)\$ | |
| Aug 28, 2014 at 13:02 | history | edited | Pimgd | CC BY-SA 3.0 |
final parameteeer
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| Aug 28, 2014 at 12:11 | history | edited | Pimgd | CC BY-SA 3.0 |
explanation of changes
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| Aug 28, 2014 at 11:43 | history | edited | Pimgd | CC BY-SA 3.0 |
better explanations
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| Aug 28, 2014 at 11:18 | comment | added | rolfl | The instructsion say: "find all the doors that will remain open after N passes.", just sayin... ;-) But your code is more compatible with the OP's code. | |
| Aug 28, 2014 at 11:17 | comment | added | Pimgd | @rolfl I gotta print the closed ones too | |
| Aug 28, 2014 at 11:16 | comment | added | rolfl |
You could simplify the logic (and make it even faster), by using for (i = 1; i <= (int)Math.sqrt(passes); i++) {...print("The door number " + (i * i) + " is Open);}
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| Aug 28, 2014 at 11:06 | history | edited | Pimgd | CC BY-SA 3.0 |
added 128 characters in body
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| Aug 28, 2014 at 10:58 | history | edited | Simon Forsberg | CC BY-SA 3.0 |
Aligned the doors
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| Aug 28, 2014 at 10:54 | history | answered | Pimgd | CC BY-SA 3.0 |