now i get this!! i thought we have to calculate the sum upto (i-1)th index.
thanx for the clarifiacation.

On Wed, Mar 14, 2012 at 3:07 PM, Dheeraj Sharma <[email protected]
> wrote:

> you have to calculate the sum of elements which are less than..that
> particular element...this is not the question of calculating cumulative sum
>
>
> On Wed, Mar 14, 2012 at 11:22 AM, sachin sabbarwal <
> [email protected]> wrote:
>
>> @gaurav popli:  how about this one??
>>
>> findsummat(int arr[],int n)
>> {
>>    int *sum ;
>>      sum =(int*)malloc(sizeof(int)*n);
>>
>> for(int i=0;i<n;i++)
>>   sum[i] = 0;
>>
>> for(int i=0;i<n;i++)
>>    sum[i] = sum[i-1] + arr[i-1];
>> //now print the sum array
>>
>> }
>>
>> it works very well....
>> plz tell me if anything is wrong with this solution.
>>
>>
>> On Tue, Mar 13, 2012 at 12:03 PM, atul anand <[email protected]>wrote:
>>
>>> @piyush : sorry dude , didnt get your algo . actually you are using
>>> different array and i get confused which array to be considered when.
>>>
>>>
>>>
>>> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor <[email protected]>wrote:
>>>
>>>> 1)First map the array numbers into the position in which they would be,
>>>> if they are sorted,for example
>>>> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
>>>> 2)Now for each number ,find the cumulative frequency of index ( = the
>>>> corresponding number in the map - 1).
>>>> 3)Output the cumulative frequency and increase the frequency  at the
>>>> position (=the corresponding number in the map) by the number itself.
>>>> Example
>>>> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
>>>> Process the numbers now,
>>>> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is
>>>> 0. so output 0
>>>>    Increase the frequency at index 2 to the number 3.
>>>> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>>>>    Increase the frequency at index 3 to the number 5.
>>>> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
>>>> position 1 by 1.
>>>> Similarly for others.
>>>>
>>>>
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>>>
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>
>
> --
> *Dheeraj Sharma*
>
>
>
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