Re: [algogeeks] Search an array of unknown length
You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.comwrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.com wrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.comwrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
@Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.com wrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.com wrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.comwrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
See at each step you are multiplying the index to be compared by 10(say), this increase is exponential. Therefore the search is exponential and complexity is log n. Base depends on the factor by which you are multiplying for the next index to be compared. Sanju :) On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek sagarpar...@gmail.comwrote: @Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.com wrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.com wrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.comwrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
hmmm ok i found a solution in which index searching is done by 2^i which is more optimal multiplication by 10 or 2 power i ??i=0,1,2,3. On Fri, Aug 19, 2011 at 11:30 PM, Sanjay Rajpal srn...@gmail.com wrote: See at each step you are multiplying the index to be compared by 10(say), this increase is exponential. Therefore the search is exponential and complexity is log n. Base depends on the factor by which you are multiplying for the next index to be compared. Sanju :) On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek sagarpar...@gmail.comwrote: @Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.com wrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.comwrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.comwrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
Well i think it depends... because range of x and 10x is more than i and 2i no doubt multiple of 10 will give us early index but then to find number in b/w indexes is more than of 2^i On Fri, Aug 19, 2011 at 11:38 PM, Sanjay Rajpal srn...@gmail.com wrote: Multiplication by 10 or 2^i , it depends. Multiplication by 10 will be faster, I think. Sanju :) On Fri, Aug 19, 2011 at 11:05 AM, sagar pareek sagarpar...@gmail.comwrote: hmmm ok i found a solution in which index searching is done by 2^i which is more optimal multiplication by 10 or 2 power i ??i=0,1,2,3. On Fri, Aug 19, 2011 at 11:30 PM, Sanjay Rajpal srn...@gmail.com wrote: See at each step you are multiplying the index to be compared by 10(say), this increase is exponential. Therefore the search is exponential and complexity is log n. Base depends on the factor by which you are multiplying for the next index to be compared. Sanju :) On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek sagarpar...@gmail.comwrote: @Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.comwrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.comwrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.com wrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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.
Re: [algogeeks] Search an array of unknown length
Thats wat I said, it depends. Searching in the interval will compensate reaching the index earlier. So both are almost equivalent. Sanju :) On Fri, Aug 19, 2011 at 11:12 AM, sagar pareek sagarpar...@gmail.comwrote: Well i think it depends... because range of x and 10x is more than i and 2i no doubt multiple of 10 will give us early index but then to find number in b/w indexes is more than of 2^i On Fri, Aug 19, 2011 at 11:38 PM, Sanjay Rajpal srn...@gmail.com wrote: Multiplication by 10 or 2^i , it depends. Multiplication by 10 will be faster, I think. Sanju :) On Fri, Aug 19, 2011 at 11:05 AM, sagar pareek sagarpar...@gmail.comwrote: hmmm ok i found a solution in which index searching is done by 2^i which is more optimal multiplication by 10 or 2 power i ??i=0,1,2,3. On Fri, Aug 19, 2011 at 11:30 PM, Sanjay Rajpal srn...@gmail.comwrote: See at each step you are multiplying the index to be compared by 10(say), this increase is exponential. Therefore the search is exponential and complexity is log n. Base depends on the factor by which you are multiplying for the next index to be compared. Sanju :) On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek sagarpar...@gmail.comwrote: @Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.comwrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.comwrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.com wrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to
Re: [algogeeks] Search an array of unknown length
:) On Fri, Aug 19, 2011 at 11:43 PM, Sanjay Rajpal srn...@gmail.com wrote: Thats wat I said, it depends. Searching in the interval will compensate reaching the index earlier. So both are almost equivalent. Sanju :) On Fri, Aug 19, 2011 at 11:12 AM, sagar pareek sagarpar...@gmail.comwrote: Well i think it depends... because range of x and 10x is more than i and 2i no doubt multiple of 10 will give us early index but then to find number in b/w indexes is more than of 2^i On Fri, Aug 19, 2011 at 11:38 PM, Sanjay Rajpal srn...@gmail.com wrote: Multiplication by 10 or 2^i , it depends. Multiplication by 10 will be faster, I think. Sanju :) On Fri, Aug 19, 2011 at 11:05 AM, sagar pareek sagarpar...@gmail.comwrote: hmmm ok i found a solution in which index searching is done by 2^i which is more optimal multiplication by 10 or 2 power i ??i=0,1,2,3. On Fri, Aug 19, 2011 at 11:30 PM, Sanjay Rajpal srn...@gmail.comwrote: See at each step you are multiplying the index to be compared by 10(say), this increase is exponential. Therefore the search is exponential and complexity is log n. Base depends on the factor by which you are multiplying for the next index to be compared. Sanju :) On Fri, Aug 19, 2011 at 10:57 AM, sagar pareek sagarpar...@gmail.comwrote: @Sanjay yeah its the very basic idea that comes in mind but is your index searching log n ? i think no !! if yes then tell me how? On Fri, Aug 19, 2011 at 11:24 PM, Sanjay Rajpal srn...@gmail.comwrote: I forgot to mention one thing, at each comparison, store the index at which we searched previously. Sanju :) On Fri, Aug 19, 2011 at 10:53 AM, Sanjay Rajpal srn...@gmail.comwrote: You can do it very easily. I assume array is sorted and contains integers. Say start at position 1, if value at that index is equal to the value to be found, return index. else if value at that index is greater than the value to be found, we got an interval to search in. else(value at that index is smaller than the value to be found) search at location 10,then 100, then 1000 till you find an interval. Once you find an interval, perform Binary Search on this and get element in O(log n). Got it ? Sanju :) On Fri, Aug 19, 2011 at 10:48 AM, sagar pareek sagarpar...@gmail.com wrote: HI, I have encountered a problem :- You have an array of *UNKNOWN *length . And you have to find an element in O(log(n)) time without using any extra space. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google
