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Topic: 4 members of sum z in hash table (Read 1243 times) |
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transgalactic
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4 members of sum z in hash table
« on: Jan 15th, 2012, 7:22am » |
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i have a set S of n members and a number Z
i need to find 4 different members which sum is Z
and the time complexity is theta(n^2)
?
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towr
wu::riddles Moderator
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Re: 4 members of sum z in hash table
« Reply #1 on: Jan 15th, 2012, 10:02am » |
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Use a hashmap mapping A*B A+B to the pair (A,B)
Then for each A*B A+B in the table find Z/(A*B) =C*D Z-(A+B) =C+D and see if A,B,C and D are distinct.
i.e. same thing as http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs; action=display;num=1325343356 but using a hash-table.
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| « Last Edit: Jan 15th, 2012, 12:35pm by towr » |
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transgalactic
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Re: 4 members of sum z in hash table
« Reply #2 on: Jan 15th, 2012, 11:59am » |
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i cant understand the meaning of
"Use a hashmap mapping A*B to the pair (A,B)"
you want me to create all the combinations of
multiplications via double loop.
then you want me o hash them
what does the function pair(A,B) does?
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towr
wu::riddles Moderator
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Re: 4 members of sum z in hash table
« Reply #3 on: Jan 15th, 2012, 12:34pm » |
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Oh wait, I seem to have misread somewhere, you want to sum, not multiply.
Well, anyway. You want distinct numbers, right. So you need to find two pairs of numbers that sum to Z: i.e. (A+B) + (C+D) = Z
By using a hashtable that maps A+B back to the pair of numbers A and B, you can iterate over the n*(n-1)/2 sums of two numbers, and find which pair of numbers give that sum, that way you can check all four are distinct.
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| « Last Edit: Jan 15th, 2012, 12:36pm by towr » |
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