@Gene - nice, correct and elegant algorithm.
On Wed, Nov 23, 2011 at 9:33 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> @Gene
>
> Your algo is also right...Just that I followed techcoders logic and coded
> the same...pair I used to map the index of the element ..But urs working
> fine too :)
>
>
> On Wed, Nov 23, 2011 at 7:04 PM, Gene <gene.ress...@gmail.com> wrote:
>
>> Sorry I forgot to initialize p. It's fixed below.
>>
>> On Nov 23, 6:59 am, Gene <gene.ress...@gmail.com> wrote:
>> > Thanks. Maybe I'm not reading correctly, but tech coder's algorithm
>> > doesn't mention anything about pairs, which are necessary to obtain
>> > O(n). This is what I meant by "almost."
>> >
>> > In reverse order, you don't need the pairs. Its simpler.
>> >
>> > In a subroutine like yours,
>> >
>> > void find_smaller_to_right(int *a, int n)
>> > {
>> > int i, in, p=0, stk[n]; // C99 var length array
>> > for (i = n - 1; i >= 0; i--) {
>> > in = a[i];
>> > while (p > 0 && stk[p - 1] >= in) p--; // pop
>> > a[i] = (p > 0) ? stk[p - 1] : 0;
>> > stk[p++] = in; // push
>> > }
>> >
>> > }
>> >
>> > On Nov 23, 5:13 am, Ankur Garg <ankurga...@gmail.com> wrote:
>> >
>> >
>> >
>> > > Solution given by tech coder is fine and is working .. I coded it and
>> its
>> > > working perfectly using stack
>> >
>> > > On Wed, Nov 23, 2011 at 2:50 PM, Gene <gene.ress...@gmail.com> wrote:
>> > > > It's a nice problem, and this solution is almost right.
>> >
>> > > > Process the input in _reverse_ order, which means we'll also
>> generate
>> > > > output in reverse order.
>> >
>> > > > The invariant is that the stack is a sorted list - highest value on
>> > > > top - of the strictly descending subsequence of elements seen so far
>> > > > in reverse.
>> >
>> > > > So when we get a new input, we want to search backward through the
>> > > > stack to find the first smaller element. This is handy however,
>> > > > because the new input also means that when we search past an
>> element,
>> > > > it's too big to maintain the invariant, so it must be popped! We
>> can
>> > > > both find the output value and update the stack at the same time:
>> >
>> > > > stack = empty
>> > > > for next input I in _reverse order_
>> > > > while stack not empty and top of stack is >= I
>> > > > pop and throw away top of stack
>> > > > if stack is empty, output is zero
>> > > > else output top of stack
>> > > > push I
>> > > > end
>> >
>> > > > Since each item is pushed and popped no more than once, this is
>> O(n).
>> >
>> > > > Here's your example:
>> >
>> > > > #include <stdio.h>
>> >
>> > > > int main(void)
>> > > > {
>> > > > int in[] = { 1, 5, 7, 6, 3, 16, 29, 2, 7 };
>> > > > int n = sizeof in / sizeof *in - 1;
>> > > > int out[100], stk[100], p = 0, i;
>> >
>> > > > for (i = n - 1; i >= 0; i--) {
>> > > > while (p && stk[p - 1] >= in[i]) p--;
>> > > > out[i] = (p > 0) ? stk[p - 1] : 0;
>> > > > stk[p++] = in[i];
>> > > > }
>> > > > for (i = 0; i < n; i++) printf(" %d", out[i]);
>> > > > printf("\n");
>> > > > return 0;
>> > > > }
>> >
>> > > > On Nov 22, 2:20 pm, Aamir Khan <ak4u2...@gmail.com> wrote:
>> > > > > On Tue, Nov 22, 2011 at 11:50 PM, tech coder <
>> techcoderonw...@gmail.com
>> > > > >wrote:
>> >
>> > > > > > here is an O(n) approach using a stack.
>> >
>> > > > > > problem can be stated as " find the 1st smaller element on the
>> right.
>> >
>> > > > > > put the first element in stack.
>> > > > > > take next element suppose "num" if this number is less than
>> elements
>> > > > > > stored in stack, pop those elements , for these pooped
>> elements num
>> > > > will
>> > > > > > be the required number.
>> > > > > > put the the element (num) in stack.
>> >
>> > > > > > repeat this.
>> >
>> > > > > > at last the elements which are in next , they will have 0
>> (valaue)
>> >
>> > > > > > @techcoder : If the numbers are not in sorted order, What
>> benefit the
>> >
>> > > > > stack would provide ? So, are you storing the numbers in sorted
>> order
>> > > > > inside the stack ?
>> >
>> > > > > I can think of this solution :
>> >
>> > > > > Maintain a stack in which the elements will be stored in sorted
>> order.
>> > > > Get
>> > > > > a new element from array and lets call this number as m. Push m
>> into the
>> > > > > stack. Now, find all elements which are <= (m-1) using binary
>> search. Pop
>> > > > > out all these elements and assign the value m in the output array.
>> > > > Elements
>> > > > > remaining at the end will have the value 0.
>> >
>> > > > > I am not sure about the complexity of this algorithm...
>> >
>> > > > > > On Wed, Nov 23, 2011 at 12:02 AM, Anup Ghatage <
>> ghat...@gmail.com>
>> > > > wrote:
>> >
>> > > > > >> I can't think of a better than O(n^2) solution for this..
>> > > > > >> Any one got anything better?
>> >
>> > > > > >> On Tue, Nov 22, 2011 at 8:23 PM, Ankuj Gupta <
>> ankuj2...@gmail.com>
>> > > > wrote:
>> >
>> > > > > >>> Input: A unsorted array of size n.
>> > > > > >>> Output: An array of size n.
>> >
>> > > > > >>> Relationship:
>> >
>> > > > > >>> > elements of input array and output array have 1:1
>> correspondence.
>> > > > > >>> > output[i] is equal to the input[j] (j>i) which is smaller
>> than
>> > > > > >>> input[i] and jth is nearest to ith ( i.e. first element which
>> is
>> > > > smaller).
>> > > > > >>> > If no such element exists for Input[i] then output[i]=0.
>> >
>> > > > > >>> Eg.
>> > > > > >>> Input: 1 5 7 6 3 16 29 2 7
>> > > > > >>> Output: 0 3 6 3 2 2 2 0 0
>> >
>> > > > > >>> --
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>> >
>> > > > > >> --
>> > > > > >> Anup Ghatage
>> >
>> > > > > >> --
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>> >
>> > > > > > --
>> > > > > > *
>> >
>> > > > > > Regards*
>> > > > > > *"The Coder"*
>> >
>> > > > > > *"Life is a Game. The more u play, the more u win, the more u
>> win , the
>> > > > > > more successfully u play"*
>> >
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>> >
>> > > > > --
>> > > > > Aamir Khan | 3rd Year | Computer Science & Engineering | IIT
>> Roorkee
>> >
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>>
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>>
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--
Nitin Garg
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