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Thanks Peyar! It worked.
One more question,for ranges of the loop of b,c,d, can't we directly do like @hrabryi has done or like you did in your previous answer of O(n^4)? I am not able to understand the use of root function here

It was a written test...sorry i did not check that while posting the question.
Can you please explain you algo in steps or give pseudo-code ?
Thanks

@Peyar : I have edited the question and provided sample input. Can you share your view how can we achieve that.Actually all the solutions with sum less than S can be included

Thank you for such awesome solution.
If I am not wrong,the time complexity of finding cycle will be O(nlogn) as for each child node it traverses upwards and in worst case O(n^2)..
I was thinking of using graph(DFS) to solve this:
- while adding edges (addEdge(a,b)), I will check two conditions ,whether it already has 2 child nodes and if the the tuple is duplicate
- after this ,DFS of graph to check if there is a cycle, whose time complexity will be O(no of vertices)

But I am not able to check the last condition : finding more than one root.
Any suggestions here how to do that?
Thanks again

Thanks for sharing !
Here you have done is split in two cases:
- If two intervals do not intersect,then either consider it or do not consider it
- If they intersect,then without considering

Please let me know if I understood your logic right.
Also , time complexity will be 2^n of recursive algo, right?

@Roger
But if all the digits of telephone number are distinct then wont the time complexity drop to O(nlogn) which is same as comparison-sort. Please correct if I am wrong

@ChrisK
Thanks for reply!
But what if the data has random numbers, and the pivot I select does not divide into half?Is there any specific way in selecting pivot?
Generally I prefer using last element as pivot