On 4/10/13 12:13 AM, rahul sharma 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...??
First of all, the array size and number of iteration both won't be N or
else the answer will always be 0.
I take the following assumption, array have K elements and number of
iteration is N.
Now, if N >= K, the return value will always be 0.
For rest, we can decompose the array following the rule of adjacent
element difference.
Solution with O(NK) time complexity follows:
int
doit (vector <int> v, int N) {
int k = (int) v.size () - 1;
if (N > k) return 0;
int c = 1;
while (N--) {
for (int i = k; i >= c; i--)
v [i] -= v [i - 1];
for (int i = 0; i < c; i++)
v [i] = 0;
c++;
}
return accumulate (v.begin (), v.end (), 0);
}
int
main ()
int a [] = { 5, 3, 8, 9, 16 };
vector <int> v (a, a + sizeof (a) / sizeof (a [0]));
assert (doit (v, 0) == 41);
assert (doit (v, 1) == 11);
assert (doit (v, 2) == 9);
assert (doit (v, 3) == -1);
assert (doit (v, 4) == 21);
assert (doit (v, 5) == 0);
return 0;
}
However, I /strongly believe/ the solution can be done in *linear
time*. To code this with quadratic time complexity is a no-brainer.
So, I took the array with K = 6 elements and decomposed.
N = 0: [a1, a2, a3, a4, a5, a6] => a1 + a2 + a3 + a4 + a4 + a6
N = 1: [0, a2 - a1, a3 - a2, a4 - a3, a5 - a4, a6 - a5] => a6 - a1
N = 2: [0, 0, a3 - 2a2 + a1, a4 - 2a3 + a2, a5 - 2a4 + a3, a6 - 2a5 +
a4] => a6 - a5 - a2 + a1 => (a6 - a5) - (a2 - a1)
N = 3: [0, 0, 0, a4 - 3a3 +3a2 - a1, a5 - 3a4 + 3a3 - a2, a6 - 3a5 + 3a4
- a3] => a6 - 2a5 +a4 - a3 + 2a2 - a1
N = 4: [0, 0, 0, 0, a5 - 4a4 + 6a3 - 4a2 + a1, a6 - 4a5 + 6a4 - 4a3 +
a2] => a6 - 3a5 + 2a4 + 2a3 - 3a2 + a1
N = 5: [0, 0, 0, 0, 0, a6 - 5a5 + 10a4 - 10a3 + 5a2 - a1] => a6 - 5a5 +
10a4 - 10a3 + 5a2 - a1
N >= 6: [0, 0, 0, 0, 0, 0] => 0
The resulting solution does show some property of Binomial coefficient
as pointed out by @Don in his hint (Pascal triangle). I suppose this
shall be the way to attack this problem.
On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma
<rahul23111...@gmail.com <mailto:rahul23111...@gmail.com>> 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 <dondod...@gmail.com
<mailto:dondod...@gmail.com>> 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 <rahul23111...@gmail.com
<mailto:rahul23111...@gmail.com>> 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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