int getsum(int *a, int n)
{
while(--n)
{
for(int i = 0; i < n; ++i)
a[i] = a[i+1] - a[i];
}
return a[0];
}
I'm not really clear about how it is intended to work. It seems that
if you start with an array of N values, each iteration reduces the
number of values by 1, so after N iterations there will be no values
left, so you could simply return 0. But I don't think that is the
intended solution.
Recursion is an elegant solution to some problems which involve
backtracking or a branching search. Using recursion to accomplish a
simple loop is wasteful and unnecessary.
For bonus points, can you determine how this problem relates to
Pascal's Triangle.
Don
On Apr 9, 2:43 pm, rahul sharma <[email protected]> wrote:
> If you have any other solution ..please post that...i thnik recursion is ok
> with base case...we need to scan again after first iteration...??
>
> On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma <[email protected]>wrote:
>
>
>
>
>
>
>
> > i forgot to add base case..can add wen 2 elemnts are there then there sum
> > is stored and we reurn from there...i m in hurry,,,sry for that,,
>
> > On Wed, Apr 10, 2013 at 12:11 AM, Don <[email protected]> wrote:
>
> >> It is O(N^2) because the inner loop takes N steps to execute and that
> >> loop will be executed N times.
>
> >> However, I would suggest not using recursion. There is no reason to
> >> not do it iteratively. Your recursive solution has no base case so it
> >> will recurse until your computer runs out of stack space, at which
> >> point it will crash.
>
> >> Don
>
> >> On Apr 9, 2:29 pm, rahul sharma <[email protected]> wrote:
> >> > A = {5, 3, 8, 9, 16}
> >> > After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
> >> > After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
> >> > Given an array, return sum after n iterations
>
> >> > my sol/
> >> > void abc(int arr[],n)
> >> > {
> >> > for(i=0;i<n;i++)
> >> > arr[i]=arr[i+1]-arr[i];
> >> > abc(arr,n-1);
>
> >> > }
>
> >> > I wana ask that the complexity is o(n) or o(n)2......as loop is
> >> executed n
> >> > times..say n is 10...so fxn is called 10 times....i.e 10 n..and
> >> ignoring n
> >> > it comes out to be...n......but if we implemeted with 2 loops then
> >> > complexity is n2 ...and both sol are taking same no of
> >> iterations...please
> >> > tell whether complexity is n or n2 for above code....if it is n2 then
> >> how???
>
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