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
>>> Google
>>> > > > > Groups
>>> > > > > > > "Algorithm Geeks" group.
>>> > > > > > > To post to this group, send email to
>>> [email protected].
>>> > > > > > > To unsubscribe from this group, send email to
>>> > > > > > > [email protected].
>>> > > > > > > For more options, visit this group at
>>> > > > > > >http://groups.google.com/group/algogeeks?hl=en.
>>> >
>>> > > > > > --
>>> > > > > > Regards
>>> > > > > > Gunjan Sharma
>>> > > > > > B.Tech IV year CSE
>>> >
>>> > > > > > Contact No- +91 9997767077
>>> >
>>> > > > > --
>>> > > > > You received this message because you are subscribed to the
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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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