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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