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So, I have this task: Let us have square matrix $A \ size\ n\ \times\ n$ for which is true: $$A(k,l)<=A(m,p)\ if \ k <=m, l<=p$$ I need to find algorithm, which finds value X in such matrix, which is not very hard, it combination of binary search, first to find row(or column), and then binary search in row(or column) which will be $~nlog(n)$. But second task is: prove that there is no algorithm which will do it for $O(n)$ time complexity that is the point where I'm stuck. I know how to calculate complexity of some algorithm, but I don't know how to prove that my algorithm is fastest.

Thank you!

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    $\begingroup$ Size $n$ meaning that it is $n \times n$? $\endgroup$ – dalastboss Mar 15 '15 at 21:06
  • $\begingroup$ Yes, I think, I'll change it to $n\ \times\ n$ $\endgroup$ – DoctorMoisha Mar 16 '15 at 19:02
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I think this is not correct. The problem can actually be solved in $O(n)$.

Consider the following algorithm: Find the matrix element by starting in the top right corner at element $A(1, n)$ and moving towards the lower left corner $A(n, 1)$. In each step return $A(k,l)$ if $X=A(k,l)$, move left (i.e. decrease $l$) if $X>A(k,l)$ and move down (i.e. increase $k$) if $A(k,l)<X$.

Too see that this works, consider the upper right submatrix with elements $A(i,j)$ for $i=1,\ldots,k$ and $j=l,\ldots,n$ for $A(k,l)=X$. When walking from $A(1,n)$ towards $A(n,1)$, we are leaving the upper right submatrix along the row or column where $X$ lies. We must pass $X$, because on the row where $X$ lies all elements on the right side of $X$ are large then $X$, so each step is a left move. In the same way if we are on one of the elements in the column where $X$ lies, each following step will be a down move.

See also here.

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