@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 > > > > > > > >>> -- > > > > > >>> You received this message because you are subscribed to the > Google > > > > > >>> Groups "Algorithm Geeks" group. > > > > > >>> To post to this group, send email to > algogeeks@googlegroups.com. > > > > > >>> To unsubscribe from this group, send email to > > > > > >>> algogeeks+unsubscr...@googlegroups.com. > > > > > >>> For more options, visit this group at > > > > > >>>http://groups.google.com/group/algogeeks?hl=en. > > > > > > > >> -- > > > > > >> Anup Ghatage > > > > > > > >> -- > > > > > >> You received this message because you are subscribed to the > Google > > > > Groups > > > > > >> "Algorithm Geeks" group. > > > > > >> To post to this group, send email to algogeeks@googlegroups.com > . > > > > > >> To unsubscribe from this group, send email to > > > > > >> algogeeks+unsubscr...@googlegroups.com. > > > > > >> For more options, visit this group at > > > > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > > > > > -- > > > > > > * > > > > > > > > Regards* > > > > > > *"The Coder"* > > > > > > > > *"Life is a Game. The more u play, the more u win, the more u > win , the > > > > > > more successfully u play"* > > > > > > > > -- > > > > > > You received this message because you are subscribed to the > Google > > > > Groups > > > > > > "Algorithm Geeks" group. > > > > > > To post to this group, send email to algogeeks@googlegroups.com. > > > > > > To unsubscribe from this group, send email to > > > > > > algogeeks+unsubscr...@googlegroups.com. > > > > > > For more options, visit this group at > > > > > >http://groups.google.com/group/algogeeks?hl=en. > > > > > > > -- > > > > > Aamir Khan | 3rd Year | Computer Science & Engineering | IIT > Roorkee > > > > > > -- > > > > You received this message because you are subscribed to the Google > Groups > > > > "Algorithm Geeks" group. > > > > To post to this group, send email to algogeeks@googlegroups.com. > > > > To unsubscribe from this group, send email to > > > > algogeeks+unsubscr...@googlegroups.com. > > > > For more options, visit this group at > > > >http://groups.google.com/group/algogeeks?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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