@Sanjay awesome solution
it won't be O(n^2) in worst case, it will be O(m+n) only

On Sat, Aug 20, 2011 at 3:22 PM, Sanjay Rajpal <[email protected]> wrote:

> Yes, thats right.
> I think we can do the following also :
>
> Lets us assume rows are sorted in increasing order.
>
> start from first row say i. Traverse the array from the end of the row
> towards the beginning till 0 occurs say at position j.
> now proceed to the next row i+1, check the value at i+1,j  if it is 0, go
> to next row i+2,j
> else if its 1, then go to left till 0 occurs and store that index of 0 and
> follow to the next row.
>
> In the worst case, it will be O(n^2), but in general its a good approach i
> guess. what do u say guys ?
>
> Average Case O(m+n) ?
>
>
> Sanju
> :)
>
>
>
> On Sat, Aug 20, 2011 at 2:47 AM, shady <[email protected]> wrote:
>
>> binary search on every row which will give solution in O(m*(logn))
>>
>>  On Sat, Aug 20, 2011 at 3:13 PM, Sanjay Rajpal <[email protected]> wrote:
>>
>>>  Sorry I forgot to mention that.
>>>
>>> Sanju
>>> :)
>>>
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