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By MODDI, history, 21 month(s) ago, In English

We have an N*N matrix, can we achieve better than transposing the matrix into an array, and then sorting it?

I did some online searching and found that quickselect eliminates the log factor, how can we extend quickselect to 2 dimensions?

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21 month(s) ago, # |
Rev. 3   Vote: I like it +38 Vote: I do not like it

Convert the matrix to an array first, then use nth_element(x,x+n/2,x+n). The overall complexity is the same as the input complexity, which has reached the lower bound.

nth_element is $$$\Theta(n)$$$.

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    21 month(s) ago, # ^ |
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    nth_element is $$$O(n) $$$, not $$$\Theta(n)$$$

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      21 month(s) ago, # ^ |
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      You're wrong, it's $$$\Theta(n)$$$, because it takes at least a linear perusal of all the elements to determine the value of the nth.

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21 month(s) ago, # |
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Just use the Thanos algorithm: among $$$n^2$$$ elements of the matrix, select random $$$n$$$ and find their median with nth_element ;-)

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    21 month(s) ago, # ^ |
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    How can we efficiently select n random elements from the matrix without iterating over the full matrix?

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    21 month(s) ago, # ^ |
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    And let's we want to extend the program, and find the smallest median among every submatrix of size K*K, how can we solve this?

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      21 month(s) ago, # ^ |
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      binary search

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        21 month(s) ago, # ^ |
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        With binary search wouldnt it be N^2*K*logK, but can we achieve N^2*K or better?

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21 month(s) ago, # |
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