nope... it is O(m+n) only :)

On Sat, Aug 20, 2011 at 3:30 PM, Shashank Gupta <[email protected]>wrote:

> In worst case it would be O(m*n)..
>
>
>
> On Sat, Aug 20, 2011 at 3:27 PM, shady <[email protected]> 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 <[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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> $hashank Gupta
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