For any query (x1, x2, y), ans1 stores answer for query(0, x1, y) and ans2 stores answer for query(0, x2, y). So our final answer becomes (ans2 - ans1) which will be storing answer for query (x1, x2, y).
Kind of inclusion and exclusion on the count of intersection.
Updated my solution with comments.
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Thanks. You had already explained your method. I asked for the comments from the main author of the post. But thanks ,now your answer code is really very simple to understand.
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Awesome approach!
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I solved it using policy based data structures (my solution here). Did anybody else use this same approach?
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I thought about it during the contest but was too lazy to study PBDS. It is also kind of an overkill for this question i guess 
Can you share some good resources to learn it?
Excellent solution. Interestingly, I thought of this approach, but could not optimize the algo/code properly.
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I mean all you have to do is type the header files and initialiser and you get a data structure that supports addition, deletion, find, and no. of elements less than given x in log n time. Basically set, but better.
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