actually, there is a way t do this in greedy... instead of taking the
local maximum value v for an index i, take the local maximum f(v) = i
+ v... this will get you the right answer...more efficient than DP?
that i'm not sure... but average case will be close to linear....
On Jan 14, 9:29 pm, SVIX <saivivekh.swaminat...@gmail.com> wrote:
> @pacific..
> the approach needs a little bit of tuning...
>
> for instance, for the set 9 8 7 6 5 4 3 2 1 1 2, per ur approach, u
> will pick 9, 8, 7, 6 etc...
>
> minimum jumps in reality is from 9 to 1 to 2.
>
> On Jan 14, 8:19 pm, pacific pacific <pacific4...@gmail.com> wrote:
>
> > At each location if the value is k ,
> > find the largest value in the next k elements and jump there.
>
> > This greedy approach works in 0(n^2) and i believe it works. If not can
> > someone give me a counter example ?
>
> > On Sat, Jan 15, 2011 at 3:30 AM, Avi Dullu <avi.du...@gmail.com> wrote:
> > > @jammy Even I felt the same, but the greedy 'algo' u suggest is actually
> > > IMHO not a greedy approach. You just take each arr[i] and jump *without
> > > deciding a locally optimal policy* . SO, if u were to *see* arr[i] and
> > > *decide* on the optimal policy I feel one would follow d same steps as in
> > > a
> > > DP solution. Its only just that the implementation would be O(n^2). Just
> > > to
> > > add, this is the greedy approach I feel:
>
> > > greedy_min_steps[n]
> > > for i = 0; i < n; i++:
> > > for (j = 0; j < input[i]; j++)
> > > greedy_min_steps[ i + j ] = min(greedy_min_step[ i + j ],
> > > greedy_min_steps[ i ] + 1)
>
> > > this is the greedy approach I build and I see this being exactly similar
> > > to
> > > my DP approach. There are instances of greedy approach based algorithms
> > > which have *optimized* DP counter parts. I feel this problem is one of
> > > them.
> > > More ideas ?
>
> > > Programmers should realize their critical importance and responsibility in
> > > a world gone digital. They are in many ways similar to the priests and
> > > monks
> > > of Europe's Dark Ages; they are the only ones with the training and
> > > insight
> > > to read and interpret the "scripture" of this age.
>
> > > On Sat, Jan 15, 2011 at 2:14 AM, Jammy <xujiayiy...@gmail.com> wrote:
>
> > >> @Avi Greedy approach doesn't work since you can't ensure the choice is
> > >> locally optimum. Consider 3,9,2,1,8,3. Using greedy alg. would give
> > >> you 3,1,8,3 while otherwise DP would give you 3,9,3.
>
> > >> On Jan 14, 6:11 am, Avi Dullu <avi.du...@gmail.com> wrote:
> > >> > I guess u got confused with the comment I wrote, I have added 2 print
> > >> > statements and now I guess it should be clear to you as to why the code
> > >> is
> > >> > O(n). The comment means that each element of the min_steps_dp will be
> > >> > ACCESSED only ONCE over the execution of the entire program. Hence the
> > >> outer
> > >> > loop still remains O(n). The next_vacat variable if u notice is always
> > >> > incremental, never reset to a previous value.
>
> > >> > #include<stdio.h>
> > >> > #include<stdlib.h>
>
> > >> > #define MAX 0x7fffffff
>
> > >> > inline int min(int a, int b) {
> > >> > return a >= b ? b : a;
>
> > >> > }
>
> > >> > int find_min_steps(int const * const input, const int n) {
> > >> > int min_steps_dp[n], i, temp, next_vacant;
> > >> > for (i = 0; i < n; min_steps_dp[i++] = MAX);
>
> > >> > min_steps_dp[0] = 0;
> > >> > next_vacant = 1; // Is the index in the array whose min_steps needs
> > >> > to
> > >> be
> > >> > updated
> > >> > // in the next iteration.
> > >> > for (i = 0; i < n && min_steps_dp[n - 1] == MAX; i++) {
> > >> > temp = i + input[i];
> > >> > if (temp >= n) {
> > >> > min_steps_dp[n - 1] = min(min_steps_dp[n - 1], min_steps_dp[i] +
> > >> 1);
> > >> > temp = n - 1;
> > >> > } else {
> > >> > printf("Updating min[%d] to %d \n", i + input[i], min_steps_dp[i]
> > >> +
> > >> > 1);
> > >> > min_steps_dp[temp] = min(min_steps_dp[temp], min_steps_dp[i] +
> > >> > 1);
> > >> > }
> > >> > if (temp > next_vacant) {
> > >> > printf("i: %d \n", i);
> > >> > for (; next_vacant < temp; next_vacant++) {
> > >> > printf("next_vacant: %d \n", next_vacant);
> > >> > min_steps_dp[next_vacant]
> > >> > = min(min_steps_dp[temp], min_steps_dp[next_vacant]);
> > >> > }
> > >> > }
> > >> > }
> > >> > for (i=0;i<n;printf("%d ",min_steps_dp[i++]));printf("\n");
> > >> > return min_steps_dp[n-1];
>
> > >> > }
>
> > >> > int main() {
> > >> > int n, *input, i;
> > >> > scanf("%d",&n);
> > >> > if ((input = (int *)malloc(n * sizeof(int))) == NULL) {
> > >> > return -1;
> > >> > }
> > >> > for (i = 0;i < n; scanf("%d",&input[i++]));
> > >> > printf("Minimum steps: %d\n",find_min_steps(input, n));
> > >> > return 0;
>
> > >> > }
>
> > >> > Programmers should realize their critical importance and responsibility
> > >> in a
> > >> > world gone digital. They are in many ways similar to the priests and
> > >> monks
> > >> > of Europe's Dark Ages; they are the only ones with the training and
> > >> insight
> > >> > to read and interpret the "scripture" of this age.
>
> > >> > On Fri, Jan 14, 2011 at 1:49 PM, Decipher <ankurseth...@gmail.com>
> > >> wrote:
> > >> > > I don't think the inner loop is executing only once . Kindly check it
> > >> for
> > >> > > this test case {1,3,5,8,9,2,6,7,6,8,9} . And try to print i in inner
> > >> loop
> > >> > > you will find that for same values of i(Outer index) inner loop is
> > >> called.
> > >> > > Its an O(n2) solution .
>
> > >> > > --
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