@ross: I couldn't get reddy's solution. Please explain.
On Sun, Jun 5, 2011 at 10:50 PM, Deepak Jha <deepak.127.0....@gmail.com>wrote:
> the below solution should work given the input array is sorted ( I am
> assuming ascending order)
> void rearrangeArray(int[] a, int[] b){
> int m = a.length;
> int n = b.length;
> int i = m - 1;
> int j = 0;
> while((i >=0) && (j <= n-1)){
> if(a[i] > b[j]){
> int temp = a[i];
> a[i] = b[j];
> b[j] = temp;
> }
> i--;
> j++;
> }
> }
>
> On Sat, Jun 4, 2011 at 2:29 PM, ross <jagadish1...@gmail.com> wrote:
>
>> Hi Rohit & all,
>> Sorry that there was a small typo in the 'n' 'm' texts.
>> The example given by me is anyway the correct one.
>> Sravan Reddy's solution worked fine.
>>
>> On Jun 4, 10:08 am, rohit <rajuljain...@gmail.com> wrote:
>> > i think solution would be like this
>> >
>> > 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.)
>> >
>> > this example is given by ROSS itself.
>> >
>> > so sravanreddy solution is right , correct me if i'm wrong.
>> >
>> > On Jun 3, 8:07 pm, bittu <shashank7andr...@gmail.com> wrote:
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > > @sravanreddy...logical bugs if A is size of n & B is size m from your
>> > > example assuming n<m so if you want smallest m elements in A then u
>> > > only capacity of n elements & didn't allocate memory so these elements
>> > > initialized by INT_MIN for m-n nodes so thatarrayA can hold m
>> > > smallest elements then what r u swapping u dude..isn't garbage
>> > > value ?? you will get at 1st step only just run it ?? in you algo
>> > > A_End=m-1(which 4th position inArraythat DNE)..?? & also you have to
>> > > free memory for m-n inarrayB as it contains n largest elements .
>> >
>> > > take
>> > > A= 1,2,3 n=3
>> > > B= 0,1,4,5,9 m=5
>> >
>> > > after allocating memory toArrayA for m-n elements A will looks
>> > > likes 1 2 3 INT_Max INT_Max
>> > > now what you wants A should contains m smallest elements & B have n
>> > > largest elements
>> > > so O/P should be A=1,2,3,1,0 & B=INT_Max,INT_Max,4,5,9 now free
>> > > memory used by 1st elements inarrayB so that A will represent M
>> > > smallest elements & B will have n Largest elements
>> >
>> > > so that above will work.
>> >
>> > > Hope I am Correct let me know if any issue with explanation
>> >
>> > > Thanks
>> > > Shashank>>"The Best Way To Escape From Theproblemis To Solve It"
>> > > CSE,BIT Mesra
>>
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-Aakash Johari
(IIIT Allahabad)
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