Most efficient way to search a sorted matrix?

Question

I have an assignment to write an algorithm (not in any particular language, just pseudo-code) that receives a matrix [size: M x N] that is sorted in a way that all of it\'s

Answer 1

in log M you can get a range of rows able to contain the target (binary search on the first value of rows, binary search on last value of rows, keep only those rows whose first <= target and last >= target) two binary searches is still O(log M)
then in O(log N) you can explore each of these rows, with again, a binary search!

that makes it O(logM x logN)
tadaaaa

Answer 2

this is wrong answer

I am really not sure if any of the answers are the optimal answers. I am going at it.

1. binary search first row, and first column and find out the row and column where "x" could be. you will get 0,j and i,0. x will be on i row or j column if x is not found in this step.
2. binary search on the row i and the column j you found in step 1.

I think the time complexity is 2* (log m + log n).

You can reduce the constant, if the input array is a square (n * n), by binary searching along the diagonal.

Answer 3

Your matrix looks like this:

a ..... b ..... c
. .     . .     .
.   1   .   2   . 
.     . .     . .
d ..... e ..... f
. .     . .     .
.   3   .   4   .
.     . .     . .
g ..... h ..... i

and has following properties:

a,c,g < i  
a,b,d < e
b,c,e < f
d,e,g < h
e,f,h < i

So value in lowest-rigth most corner (eg. i) is always the biggest in whole matrix and this property is recursive if you divide matrix into 4 equal pieces.

So we could try to use binary search:

1. probe for value,
2. divide into pieces,
3. choose correct piece (somehow),
4. goto 1 with new piece.

Hence algorithm could look like this:

input: X - value to be searched
until found
 divide matrix into 4 equal pieces
 get e,f,h,i as shown on picture
 if (e or f or h or i) equals X then 
   return found
 if X < e then quarter := 1
 if X < f then quarter := 2
 if X < h then quarter := 3
 if X < i then quarter := 4
 if no quarter assigned then 
    return not_found
 make smaller matrix from chosen quarter 

This looks for me like a O(log n) where n is number of elements in matrix. It is kind of binary search but in two dimensions. I cannot prove it formally but resembles typical binary search.