@Ashish: your solution is not O(N). I dont think you are taking advantage of
the statement "( A and B need not be sorted in the end)"
@sravanreddy: excellent solution.
On Thu, Jun 2, 2011 at 7:46 PM, Ashish Goel <ashg...@gmail.com> wrote:
>
> int i=lenA-1;
> int j=lenB-1;
>
> while (j>=0)
> {
> if (A[i] >B[j]) {swap(A[i] ,B[j]); sort(A); }
> j--;
> }
>
>
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Sat, May 28, 2011 at 11:09 PM, ross <jagadish1...@gmail.com> wrote:
>
>> Hi all,
>> Given 2 sorted arrays: A and B each holding n and m elements
>> respectively,.
>> Hence, total no. of elements = m+n
>> Give an algorithm to place the smallest 'm' elements(out of the m+n
>> total available) in A and the largest 'n' elements in B. ( A and B
>> need not be sorted in the end)
>>
>> eg:
>> A : 1 2 3 B: 0 1.5 4 5 9
>>
>> Output:
>> A can contain any combination of nos 0,1,1.5
>> and B should contain 2 3 4 5 9 (in any order.)
>>
>> Constraints: No extra space. Linear Time preferred
>>
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