peeyush, take eg: n=6 array values: 10 20 30 40 50 50 in worst case, while loop which can increment 'i' can go upto n-1 and for loop (for 'j') every n-1 time check upto n times so total it becomes (n-1)*n= O(n*n).
like this i think u can observe now.. On 8/18/07, Peeyush Bishnoi <[EMAIL PROTECTED]> wrote: > > Thanxs for giving feedback... :-) > Can you please explain how worst case time complexity is O(n*n) of this > solution. Means how u determine this. Plz explain.... > > --- > Peeyush Bishnoi > > On 8/18/07, Vaibhav Jain <[EMAIL PROTECTED]> wrote: > > > > hi peeyush, > > > > ur solution is nice, it is brute force method and space complexity is > > constant here > > but ur solution contains worst case time complexity O(n*n) > > and we want O(n) solution. So ur solution is not required solution. > > > > > > On 8/18/07, Peeyush Bishnoi < [EMAIL PROTECTED]> wrote: > > > > > > Hello All , > > > > > > I am thinking this solution offered by me is some what accurate with > > > constant space . Just put ur feed back on this. > > > > > > If you have any query ask me. > > > > > > int main(){ > > > int a[]={1,2,2,3}; > > > int count=sizeof(a)/sizeof(a[0]); > > > printf("No.of elemenst:%d\n",count); > > > fun(a,count); > > > return 0; > > > } > > > > > > fun(int a[],int count){ > > > int i=0,j; > > > while(i<count){ > > > for(j=0;(j<i && (a[i]!=a[j]));j++); > > > if((j<i)&& (a[i]==a[j])){ > > > printf("No. is repeated:%d\n",a[i]); > > > }else{ > > > } > > > i++; > > > } > > > } > > > > > > --- > > > Peeyush Bishnoi > > > > > > On 8/18/07, Dondi Imperial < [EMAIL PROTECTED] > wrote: > > > > > > > > > > > > hi, > > > > > > > > actually in mine space complexity is O(n) in _all_ cases. :). Out > > > > of > > > > curiousity, will your solution work when not all the numbers in the > > > > range are present in the array? > > > > > > > > Thanks, > > > > > > > > Dondi > > > > > > > > On 8/17/07, Vaibhav Jain < [EMAIL PROTECTED] > wrote: > > > > > hello Dondi, > > > > > > > > > > in ur solution, space complexity will be O(n) in worst case. > > > > > but in my solution it will constant space with linear complexity. > > > > > > > > > > now think abt how to prove it if range is not known for numbers > > > > > then can we achieve it or not? > > > > > if not then prove it....??? > > > > > > > > > > > > > > > > > > > > > > > > > On 8/17/07, Dondi Imperial < [EMAIL PROTECTED]> wrote: > > > > > > > > > > > > if you know the range of the numbers don't you just have to > > > > create and > > > > > > array (of length k in your example) then iterate over the array > > > > and > > > > > > increment the corresponding element in the other array. > > > > > > > > > > > > Ie, > > > > > > > > > > > > int[] arrayValues = some array of a known range > > > > > > int[] arrayLookup = int[min_in_range - max_in_range + 1] > > > > > > > > > > > > foreach(i in arrayValues) > > > > > > if(arrayLookup[i] > 0) then > > > > > > found > > > > > > else > > > > > > arrayLookup[i]++ > > > > > > > > > > > > Of course range could be prohibitively large (still constant > > > > though if > > > > > > you know the range before hand). > > > > > > > > > > > > > > > > > > > > > > > > On 8/17/07, Vaibhav Jain < [EMAIL PROTECTED]> wrote: > > > > > > > if u know the range of values stored in array then > > > > > > > let me assume values 1 to k then u can calculate sum of > > > > numbers stored > > > > > in > > > > > > > array in O(n) complexity. > > > > > > > after that apply formula > > > > > > > > > > > > > > duplicate value= [k*(k+1)]/2 - sum of numbers stored in array > > > > > > > > > > > > > > it will take O(1) constant time so total complexity becomes > > > > only O(n). > > > > > > > > > > > > > > it can be one solution to your problem but if the range is > > > > unknown for > > > > > > > values then > > > > > > > is there any solution to come in O(n)??? > > > > > > > > > > > > > > > > > > > > > > > > > > > > On 8/16/07, dsha < [EMAIL PROTECTED]> wrote: > > > > > > > > > > > > > > > > Hi there, > > > > > > > > > > > > > > > > I'm interested in the following problem: there is an array > > > > of integers > > > > > > > > that contains each element only once except for one element > > > > that > > > > > > > > occurs exactly twice. Is there a way to find this element > > > > faster than > > > > > > > > O(n*log n) and with constant extra memory? If no, how can I > > > > prove it? > > > > > > > > > > > > > > > > Thanks in advance for ideas. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > > > > > > > Vaibhav Jain > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > > > > > Vaibhav Jain > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > > Vaibhav Jain > > > > > > > > > > > > -- Vaibhav Jain --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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