@bhavana : Explain..!!
as far as i get you is that  it would  be same as implementing map ...!!
isn't ??

On Fri, May 27, 2011 at 2:16 PM, bhavana <[email protected]> wrote:

> can be solved easily if the elements of the array lie in a limited range
> which can b known beforehand...!!
>
>
> On Fri, May 27, 2011 at 2:10 PM, Aakash Johari <[email protected]>wrote:
>
>> yes, you are right. map insertion takes O(log n) time. so if you know the
>> upper bound of N, you can simply map the existence/non-existence of any
>> particular element in an array. that will be in constant time (for query
>> purposes) and O(n) time for preprocessing.
>>
>>
>> On Fri, May 27, 2011 at 1:29 AM, sukhmeet singh 
>> <[email protected]>wrote:
>>
>>> @Dave nd @Akash can u explain a bit more.. I didn't get what u say..
>>> Inserting in a map takes O(log n)  time !!
>>>
>>> On Fri, May 20, 2011 at 8:35 PM, Aakash Johari <[email protected]>wrote:
>>>
>>>> @Dave: This is what is still a doubt to me. I have searched but couldn't
>>>> get the info regarding this.
>>>>
>>>>
>>>> On Fri, May 20, 2011 at 8:01 AM, Dave <[email protected]> wrote:
>>>>
>>>>> @Aakash: And tell me how map works. Is making an entry O(1) regardless
>>>>> of the value of the entry? For example, is it O(n) to map the sequence
>>>>> 1, 4, 9, 16, 25, ..., n^2?
>>>>>
>>>>> Dave
>>>>>
>>>>> On May 20, 9:39 am, Aakash Johari <[email protected]> wrote:
>>>>> > @Dave: I got you. I will have to check before pushing the element in
>>>>> map.
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> >
>>>>> > On Fri, May 20, 2011 at 7:30 AM, Dave <[email protected]>
>>>>> wrote:
>>>>> > > @Aakash: Yeah, but try the same array with sum = 6 and see what
>>>>> > > happens.
>>>>> >
>>>>> > > Dave
>>>>> >
>>>>> > > On May 20, 9:04 am, Aakash Johari <[email protected]> wrote:
>>>>> > > > int main()
>>>>> > > > {
>>>>> > > >         int a[10] = {5, 3, 10, 9, 8, 23, 11, 4, 12, 6};
>>>>> > > >         int i;
>>>>> > > >         int sum;
>>>>> > > >         int flag = 0;
>>>>> >
>>>>> > > >         map<int, int> m;
>>>>> >
>>>>> > > >         for ( i = 0; i < 10; i++ ) {
>>>>> > > >                 m[a[i]] = 1;
>>>>> > > >         }
>>>>> >
>>>>> > > >         sum = 13;
>>>>> >
>>>>> > > >         for ( i = 0; i < 10; i++ ) {
>>>>> > > >                 if ( m[sum - a[i]] == 1 ) {
>>>>> > > >                         flag = 1;
>>>>> > > >                         break;
>>>>> > > >                 }
>>>>> > > >         }
>>>>> >
>>>>> > > >         if ( flag == 1 )
>>>>> > > >                 cout << a[i] << " " << sum - a[i] << endl;
>>>>> >
>>>>> > > >         return 0;
>>>>> >
>>>>> > > > }
>>>>> > > > On Fri, May 20, 2011 at 7:01 AM, hari <[email protected]>
>>>>> wrote:
>>>>> > > > > We can sort using STL sort function in main() before function
>>>>> call of
>>>>> > > > > arraysum().
>>>>> >
>>>>> > > > > On May 20, 6:49 am, Gunjan Sharma <[email protected]>
>>>>> wrote:
>>>>> > > > > > First of all there is an infinite loop in this code....
>>>>> > > > > > Secondly it works only for sorted array.
>>>>> >
>>>>> > > > > > On Fri, May 20, 2011 at 7:16 PM, hari <[email protected]>
>>>>> wrote:
>>>>> > > > > > > In while loop have i,j which points first and last index of
>>>>> array.
>>>>> > > In
>>>>> > > > > > > while loop, Check the sum of a[i],a[j], If sum<k,increment
>>>>> i or
>>>>> > > else
>>>>> > > > > > > decrement j. Run the while loop till i<j..
>>>>> >
>>>>> > > > > > > CODE:
>>>>> >
>>>>> > > > > > > int arraysum(int a[], int k, int i, int j)
>>>>> > > > > > > while(i<j)
>>>>> > > > > > > {
>>>>> > > > > > >  int p=0;
>>>>> > > > > > >  int b[10];     //to store index of selected nos
>>>>> > > > > > >  sum=a[i]+a[j];
>>>>> > > > > > >  if (sum==k)
>>>>> > > > > > >  {
>>>>> > > > > > >  b[p++]=i;b[p++]=j;
>>>>> > > > > > >  }
>>>>> > > > > > >  elseif(sum<k)
>>>>> > > > > > >  i++;
>>>>> > > > > > >  else(sum>k)
>>>>> > > > > > >  j++;
>>>>> > > > > > >  return b;
>>>>> > > > > > > }
>>>>> >
>>>>> > > > > > > On May 20, 4:38 am, amit <[email protected]> wrote:
>>>>> > > > > > > > given an array of integers, and an integer k, find out
>>>>> two
>>>>> > > elements
>>>>> > > > > > > > from the array whose sum is k in O(n) time. if no such
>>>>> element
>>>>> > > exists
>>>>> > > > > > > > output none.
>>>>> >
>>>>> > > > > > > --
>>>>> > > > > > > You received this message because you are subscribed to the
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>>>>> > > > > > > "Algorithm Geeks" group.
>>>>> > > > > > > To post to this group, send email to
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>>>>> > > > > > > To unsubscribe from this group, send email to
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>>>>> > > > > > > For more options, visit this group at
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>>>>> >
>>>>> > > > > > --
>>>>> > > > > > Regards
>>>>> > > > > > Gunjan Sharma
>>>>> > > > > > B.Tech IV year CSE
>>>>> >
>>>>> > > > > > Contact No- +91 9997767077
>>>>> >
>>>>> > > > > --
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>>>>> >
>>>>> > > > --
>>>>> > > > -Aakash Johari
>>>>> > > > (IIIT Allahabad)- Hide quoted text -
>>>>> >
>>>>> > > > - Show quoted text -
>>>>> >
>>>>> > > --
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>>>>> >
>>>>> > --
>>>>> > -Aakash Johari
>>>>> > (IIIT Allahabad)- Hide quoted text -
>>>>> >
>>>>> > - Show quoted text -
>>>>>
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>>>>>
>>>>
>>>>
>>>> --
>>>> -Aakash Johari
>>>> (IIIT Allahabad)
>>>>
>>>>
>>>>
>>>>
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>>
>>
>>
>> --
>> -Aakash Johari
>> (IIIT Allahabad)
>>
>>
>>
>>
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