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    <p>I'll give the solution in general words without programming it.</p> <p>Let's denote segments as s<sub>1</sub>, s<sub>2</sub>, ..., s<sub>n</sub>. Their beginnings as b<sub>1</sub>, b<sub>2</sub>,... b<sub>n</sub>, and their ends as e<sub>1</sub>, e<sub>2</sub>,... e<sub>n</sub>.</p> <p>Sort segments by their beginnings, so b<sub>1</sub>&lt; b<sub>2</sub>&lt;...&lt; b<sub>n</sub>. It is enough to check them if the condition of no three segments covering a point holds. We will be doing so in the order from b<sub>1</sub> to b<sub>n</sub>. So, start with b<sub>1</sub>, move to the next point, and so on one by one, until at some point b<sub>i</sub> there are three segments covering it. These will be the segment s<sub>i</sub> and two others, let's say s<sub>j</sub> and s<sub>k</sub>. Of those three segments delete the one with the maximum end point, i.e. max{e<sub>i</sub>, e<sub>j</sub>, e<sub>k</sub>}. Move on to the beginning of the next segment (b<sub>i+1</sub>). When we reach b<sub>n</sub> the process is done. All the segments that are left constitute the largest subset of segments such that no three segments share a common point.</p> <p>Why this will be the maximal subset. Let's say our solution is S (the set of segments). Suppose there is an optimal solution S*. Again, sort the segments in S and S* by the coordinate of their beginnings. Now, we will be going through the segments in S and in S* and comparing their end points. By the construction of S for any k<sup>th</sup> segment in S its end coordinate is smaller than the end coordinate of k<sup>th</sup> segment in S* (e<sub>k</sub>&lt;=e<sub>k</sub><em>). Therefore, the number of segments in S is not less than in S</em> (moving in S* we're always outrunning S).</p> <p>If this is not convincing enough, try to think about a simpler problem at first, where no two segments can overlap. The solution is the same, but it's much more intuitive to see why it gives the right answer.</p>
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    1. POfinding largest subset of intervals
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    1. COThe solution is likely correct but the proof has holes
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    2. COok, correct me if i am wrong but lets say if I have these segments (1-2),(1-3),(1-10),(3-5),(3-6), then according to this algorithm the solution should be (1-2),(1-3),(3-5) but a better solution could be (1-10),(1-2),(3-5)(3-6)
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    3. CO@user352951 in your example (1-10), (3-5) and (3-6) overlap
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