13.3. One Dimensional Case

So, let us see first the same problem if we had only one dimensional points. If we have one dimensional points then all these points lie along the line, which we can assume this the x axis. So, we have a bunch of points and we want to find the closest point.

In this chunk, we introduce the scenario where we are working with points that only exist in one dimension. Instead of being spread out in a two-dimensional plane, these points are aligned along a single line, often represented as the x-axis. This simplification makes it easier to visualize and calculate distances between points since we only need to consider horizontal distances.

Imagine you're at a park on a straight walking path, and you want to find the closest friend standing along that path. Instead of looking around in all directions, you simply check who is nearest to you along that path.

What we can do is we can first sort them, so that we have the points in increasing order of x coordinate and then it is easy to see that the distances that we need are the distances between two adjacent points. Because, if I look at a point, the nearest point if either the one on it is left and one on it is right.

To find the closest pair of points, the first step is sorting the points based on their x-coordinates. By doing this, we can quickly identify that the closest distance will occur between adjacent points in this sorted list. The leftmost point's closest pair can only be either the point immediately to its left or right, making the problem simpler.

Think about lining up books on a shelf by height. Once sorted, you only need to check the books next to each other to find the two that are the shortest, rather than comparing every book against every other book.

I just need to scan this x 2 minus x 1 distance, then x 3 minus x 2, so I just need to scan these n minus 1 distances and then keep track of the smallest gap between these two points and that would give me the smallest distance among the overall pair of overall n points.

Once the points are sorted, the next step is to calculate the distances between each pair of adjacent points. By scanning through the sorted list, we only need to calculate n-1 distances (one for each adjacent pair). As we keep track of the smallest distance found, we can determine which two points are closest to each other efficiently.

Imagine checking the heights of your friends standing in a straight line. Instead of measuring the height difference between everyone, you only measure the differences between pairs of friends standing next to each other in line.

So, here the algorithm is n log n, because it takes n log n times to sort the points in x coordinate, after that finding the minimum is actually very easy.

The entire process of finding the closest pair in this one-dimensional case has a time complexity of O(n log n). This is primarily due to the sorting step, which is the most time-consuming part of the algorithm. Once sorted, locating the closest pair is straightforward and can be accomplished in linear time.

Think about sorting a deck of cards. The time it takes to sort them is greater than just grabbing the two top cards and seeing which one is higher. The sorting takes effort, but once they're sorted, the next task is quick.

So in one dimension this problem is very easy to solve, the challenge is to solve it in two dimensions.

In summary, when dealing with one-dimensional points, the problem simplifies significantly. The key steps involve sorting the points and then calculating distances between adjacent points. However, transitioning to two dimensions introduces added complexity that requires a different approach to handle potential points across both dimensions.

Finding the closest store when you only have to drive down one road is easier than navigating through a busy city with many streets and intersections. The principles are the same, but the navigation becomes more complex in a two-dimensional space.

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