For the series like 2,4,3,9,4,16,5,25 ur algo runs in o(n*n/2) =o(n^2)
On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin wrote:
1)Find product of the array and store it in say prod o(n) and o(1)
2)now traverse the array and check if
static int i;
tag:
while(in)
if( prod
If we can retrieve ith prime efficiently, we can do the following...
1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
3.repeat it till end
On Thursday, 12 July 2012 10:55:02
Could u please tell me the way to find number repeated odd number of time
using X-oR..??
On Sunday, 27 November 2011 00:49:59 UTC+5:30, Gene wrote:
Isn't this overkill? If you're already using a set, just check the set
before you insert each new element, and you'll discover the
duplicates:
