Assume a matrix of integers they are sorted in boh row and column vice .. how do u find a given number from the matrix in a optimal way?

Question

Sigiloso · Accepted Answer

Let the matrix is n*m matrix. Then O(n log m) solution is trivial (binary search in each row). There is a easy O(n+m) solution too. The idea is to start from upper right corner (mat[0][m-1]) and traverse toward lower left corner (mat[n-1][0]). On the way check each entry and depending on whether larger go left or down. If there is a solution you will find it on the way. Or you will arrive to a point where you can no longer move without going out of the matrix. Either way you will check at most O(n+m) entries thus the solution in O(n+m).

Ethelbert · Answer

What if you did a binary search on the diagonal....

dp · Answer

I think it could be done even better than in O(n+m). Instead of starting at the upper right corner do a binary search on last column and find the biggest element that is still smaller than the given number. Say it's gonna be A[i, m-1]. Now we could throw away all rows up to an including i (since A[i, m-1] is larger than all of these elements) and the last column. Repeat everything for a smaller matrix of size (n - i, m - 1);

Me · Answer

Anonymous, consider the matrix and you are searching for 9. Using this new algorithm we would look at rows 2 and 3 since 8 < 9 < 10 and completely miss 9 in the last row.
(1   3    5      7)
(2   3    6      8)
(3   5    8    10)
(6   9   11   12)

Ethelbert · Answer

dp, 
what do you do if the first element in the last column is already larger than the target?  I.e. there may be no number that is smaller that the target.  You could throw away that column, but that's it, I think.  If so, the worst case becomes O(m).

Sigiloso · Answer

Ok writing this 3rd time , no one will see this but for the sake of it,

Using radix sort 
Take 10 buckets and enter all numbers respectively.
For example all numbers unit digit with 9 will in 9 number bucket.
Just divide the given number (input) and into parts , using mod operator.
And using the divided number"s unit digit find that index in the bucket.

Tada .
Radix sort way the. Linear search way to find the number

Complexity O ( n  )

Phew

matrix · Answer

How about this.
for m columns - do binary range_search.
As in, start with m/2  th column, if the number is in this range...col(0)  O (log(m+n))

matrix. · Answer

oops, there is a catch....
If the number is in a range, it can be in that column, or any of the columns towards left of it....and in some cases, if its in the range.. it can still be on any columns on right.....
need to tweak a bit more.

Sigiloso · Answer

A better solution than O[log(mn)] is unlikely. An mxn array can be laid in memory as a m*n one dimensional array at its base form. So a matrix sorted by row and column simply means this mxn one dimensional array is sorted. A binary search will get you the result in O[log(mn)], that is the fastest solution available.

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