@aman: wat u dint get???

On Fri, Jul 8, 2011 at 6:34 PM, Aman Goyal <[email protected]> wrote:

> i dint get you..
>
> one loop to access the first array elements and compare with second array,
> and one logn (for) loop to binary search the second array , thts it..
> O(mlogn) is what i am able to understand.
>
> On Fri, Jul 8, 2011 at 5:52 PM, Apoorve Mohan <[email protected]>wrote:
>
>> @aman:
>>
>> Let size of *first array be m* and that of the *second array be* *n*.
>>
>> For m elements in first array you perform binary search therefore time
>> O(mlogn)
>>
>> And for those some elements of the first array you perform shifting in
>> array two...in the worst case for all the elements of first array
>> you might have to perform shifting in second array and also you might just
>> have to shift all the present (n-1) elements each time...so again in worst
>> case this whole procedure will take O(mn) time....
>>
>> so total coplexity of your idea is: O(mlogn) + O(mn)
>>
>> And if m is of the O(n) then this will take O(n^2) time in worst case.
>>
>>
>> On Fri, Jul 8, 2011 at 2:40 PM, Aman Goyal <[email protected]>wrote:
>>
>>> Algo:
>>>
>>>
>>> 1 3 77 78 90
>>> 2 5 79 81
>>>
>>> compare 1 &2 =1
>>> compare 3 &2 =2 and call binary search on 2nd array widot 2 to identify a
>>> proper position for 3 and place it there.
>>> now arrays
>>>
>>> 1 2 77 78 90
>>> 3 5 79 81
>>> 3 and 77= swap + binary
>>>
>>> compare 3 and 77, swap them
>>> find position of 77 in second array and place there. using binary search
>>>
>>> 1 2 3 78 90
>>> 5 77 79 81
>>> 78 and 5 = swap + binary search
>>>
>>> 1 2 3 5 90
>>> 77 78 79 81
>>>
>>> 90 and 77= swap+ binary
>>>
>>>
>>> 1 2 3 5 77
>>> 78 79 81 90
>>>
>>> ans found
>>>
>>> O(nlogn)
>>> binary search is O(logn) .
>>>
>>>
>>> On Fri, Jul 8, 2011 at 8:29 AM, durgesh kumar <[email protected]>wrote:
>>>
>>>>
>>>> @dumanshu
>>>>
>>>> > ok ! i got a O(n lgn) finally
>>>> > i don know exact complexity
>>>> > Let N = size of first array
>>>> > Find the first N smallest elements using one pointer in each array
>>>> > now swap the list of elements  from index 0 to second-pointer in
>>>> > second array to first array
>>>> > with first_poiner+1 to N in first Array
>>>> > now,after swapnig we need to sort both array
>>>>
>>>>
>>>>
>>>>
>>>> so complexity= n + n log n+ m log m (n is the size of of first array and
>>>> m is the size of second array)
>>>>  .
>>>> . . O(n) = (n log n ) or (m log m)
>>>> thanks
>>>> Durgesh
>>>>
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>>>
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>>
>>
>>
>> --
>> regards
>>
>> Apoorve Mohan
>>
>>
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-- 
regards

Apoorve Mohan

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