In worst case it would be O(m*n)..

On Sat, Aug 20, 2011 at 3:27 PM, shady <sinv...@gmail.com> wrote:

> @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 <srn...@gmail.com> 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 <sinv...@gmail.com> 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 <srn...@gmail.com>wrote:
>>>
>>>>  Sorry I forgot to mention that.
>>>>
>>>> Sanju
>>>> :)
>>>>
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