My bad.... but it can be made recursive :) On Aug 9, 8:17 pm, Dave <[email protected]> wrote: > @Ankuj: Yeah, but he asked for it to be recursive. Yours is iterative. > > Dave > > On Aug 9, 9:56 am, Ankuj Gupta <[email protected]> wrote: > > > > > > > > > we can do it in logn by using binary search approach found > > n is the number whose square root has to be > > > if(n==1) > > return 1; > > if(n==0) > > return 0; > > int low=0,high=n/2,mid,temp; > > while(1) > > { > > mid = (low+high)/2; > > temp = mid*mid; > > if(temp==n) > > return 1; > > else if(temp <n) > > low = mid+1; > > else > > high = mid-1; > > if(low == mid || high == mid) > > return 0; > > } > > at the end high will have the required square root if not perfect > > square or if perfect square mid will have the required answer
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