@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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