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Reservoir Sampling | GeeksforGeeks
It can be solved in O(n) time. The solution also suits well for input in the form of stream. The idea is similar to this (Link->http://ift.tt/Uyxg2Z) post. Following are the steps.
1) Create an array reservoir[0..k-1] and copy first k items of stream[] to it.
2) Now one by one consider all items from (k+1)th item to nth item.
…a) Generate a random number from 0 to i where i is index of current item in stream[]. Let the generated random number is j.
…b) If j is in range 0 to k-1, replace reservoir[j] with arr[i]
http://ift.tt/MxNpkB
http://ift.tt/Uyxg35
It can be solved in O(n) time. The solution also suits well for input in the form of stream. The idea is similar to this (Link->http://ift.tt/Uyxg2Z) post. Following are the steps.
1) Create an array reservoir[0..k-1] and copy first k items of stream[] to it.
2) Now one by one consider all items from (k+1)th item to nth item.
…a) Generate a random number from 0 to i where i is index of current item in stream[]. Let the generated random number is j.
…b) If j is in range 0 to k-1, replace reservoir[j] with arr[i]
http://ift.tt/MxNpkB
http://ift.tt/Uyxg35
from Public RSS-Feed of Jeffery yuan. Created with the PIXELMECHANICS 'GPlusRSS-Webtool' at http://gplusrss.com http://ift.tt/1roNpmw
via LifeLong Community
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