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Finding duplicates in O(n) time and O(1) space
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This is what I came up with, which doesn't require the additional sign bit:
for i := 0 to n - 1
while A[A[i]] != A[i]
swap(A[i], A[A[i]])
end while
end for
for i := 0 to n - 1
if A[i] != i then
print A[i]
end if
end for
The first loop permutes the array so t...
What is the purpose of the word 'self'?
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edited Apr 27 '10 at 23:01
answered Apr 25 '10 at 20:25
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Get notified when UITableView has finished asking for data?
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answered Jun 17 '10 at 9:01
Eric MORANDEric MORAND
5,92533 gold badges2323 silver badges3131 bronze badges
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Multiple “order by” in LINQ
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10
Yes, I concluded that too hastily based on 3.5 not being in the version dropdown in the documentation page; I should have looked all the wa...
How are booleans formatted in Strings in Python?
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answered Feb 13 '10 at 22:07
danbendanben
70.8k1818 gold badges113113 silver badges140140 bronze badges
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How to make a Python script run like a service or daemon in Linux
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– John La Rooy
Oct 21 '09 at 21:10
What happen does use a loop without termination in a Python program and then re...
How to set custom header in Volley Request
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ccpizza
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answered Jun 11 '13 at 17:43
CommonsWareCommonsWare
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How do you create different variable names while in a loop? [duplicate]
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10 Answers
10
Active
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What is the optimal Jewish toenail cutting algorithm?
...like your original algorithm best.
Since 14 out of 120 permutations work, 106 out of 120 do not. So each check has a 106/120 chance of failing.
That means the expected number of failures is:
1*(106/120) + 2*(106/120)^2 + 3*(106/120)^3 + ...
Not too hard to sum this infinite series:
S = ...
Remove not alphanumeric characters from string
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10
This doesn't remove underscores.
– kylex
Feb 3 '13 at 4:32
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