Editing Persistent data structure (section)

====Naïve method====
We start with a vertical line segment that starts off at infinity and we sweep the line segments from the left to the right. We take a pause every time we encounter an end point of these segments. The vertical lines split the plane into vertical strips. If there are {{mvar|n}} line segments then we can get <math>2\cdot n+1</math> vertical strips since each segment has {{val|2}} end points. No segment begins and ends in the strip. Every segment either it doesn't touch the strip or it completely crosses it. We can think of the segments as some objects that are in some sorted order from top to bottom. What we care about is where the point that we are looking at fits in this order. We sort the endpoints of the segments by their {{mvar|x}} coordinate. For each strip <math>s_{i}</math>, we store the subset segments that cross <math>s_{i}</math> in a dictionary. When the vertical line sweeps the line segments, whenever it passes over the left endpoint of a segment then we add it to the dictionary. When it passes through the right endpoint of the segment, we remove it from the dictionary. At every endpoint, we save a copy of the dictionary and we store all the copies sorted by the {{mvar|x}} coordinates. Thus we have a data structure that can answer any query. In order to find the segment above a point {{mvar|p}}, we can look at the {{mvar|x}} coordinate of {{mvar|p}} to know which copy or strip it belongs to. Then we can look at the {{mvar|y}} coordinate to find the segment above it. Thus we need two binary searches, one for the {{mvar|x}} coordinate to find the strip or the copy, and another for the {{mvar|y}} coordinate to find the segment above it. Thus the query time takes <math>O(\log(n))</math>. In this data structure, the space is the issue since if we assume that we have the segments structured in a way such that every segment starts before the end of any other segment, then the space required for the structure to be built using the naïve method would be <math>O(n^{2})</math>. Let us see how we can build another persistent data structure with the same query time but with a better space.