answer will be $O(m+n) .$
if we search one by one element which will take $O(m*n)$.but we can do better because every row and column is sorted in increasing order ,we can take advantage of that property.
step 1: search from right most element ( assume p) of array ,which is greater than every other element of row.we can move now left or right according to the element (assume element x is being searched)
step 2: take loop -if p = x then return position of p
if p>x then go left else go down.
for example -$\begin{bmatrix} 10 & 20 & 30 &40 \\ 11& 22& 33&44 \\ 21 & 34& 38& 50\\ 31 & 39 & 45& 60 \end{bmatrix}$ soppose we are searching for 34.so we will start from 40 and 40>34 so go left ,now we have 30 which less than 34 so go down ,now we have 33 which is less than 34 so go down,now we have 38 which is greater than 34 so go left hence we got our element 34.so in $ O(m+n) $ we can search our element.
