true did not read the question properly...

Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652


On Fri, Jun 3, 2011 at 7:19 PM, Kunal Patil <[email protected]> wrote:

> @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 <[email protected]> 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 <[email protected]> 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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