CareerCup

Why need sort?

No sort requied.
(x1, y1)-->(x2, y2)-->(x3, y3) ....->(xn, yn)
First iteration, only check x, elimated points which x is not falling into [x-5, x+5]
Second iteration, only check y, eliated potins which y is not falling into [y-5, y+5]
For the left points, calc sqrt((x-xi)*(x-xi) + (y-yi)*(y-yi))], if the reslut <=5, keep the point, otherwise, remove it.

No sort requied.
(x1, y1)-->(x2, y2)-->(x3, y3) ....->(xn, yn)
First iteration, only check x, elimated points which x is not falling into [x-5, x+5]
Second iteration, only check y, eliated potins which y is not falling into [y-5, y+5]
For the left points, calc sqrt((x-xi)*(x-xi) + (y-yi)*(y-yi))], if the reslut <=5, keep the point, otherwise, remove it.

fix the given point.
and draw a circle whose center=(given point) and r=5.
now check those points which lies with in the circle.
that will be the ans.

If you have very few points ( i.e: 5) sorting solution would be bad than simply directly checking the distance of all points. Same applies for the circle solution. Its a very debatable question and I think it is not appropriate for a phone interview. But it would be for a on person interview.

If you have very few points ( i.e: 5) sorting solution would be bad than simply directly checking the distance of all points. Same applies for the circle solution. Its a very debatable question and I think it is not appropriate for a phone interview. But it would be for a on person interview.

i suppose points are given as Point(x, y). Now have a loop

while (n--)
{
if (abs (points[n].x - xx) == 5 && abs (points[n].y - yy) == 5)
{
cout << points[n].x << points[n].y;
}
}

// xx, yy are give points
// points is an array holding points.

complexity: o(n)

Sort x coordinates using counting sort (O(n+k)).

Than for the given point and range find the distance of each element in that x range.

O(2n+k) (k is the range)