gXa's blog

Query solving in most optimised way
By gXa, history, 9 years ago, In English

Initially the array contain all 1s.

There are two type of operation:

1 A: update arr[A] = 0.

2 A: Find index of Ath 1 in the array.

Number of elements, 1<=N<=(1e6)

Number of queries, 1<=Q<=(1e6)

I tried tree statistic. However, it didn't pass.

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9 years ago, hide # |
 
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build a segment tree wich in each node you save the number of 1's in [l, r)

update is simple.

and for answering a query in the query function check if A is smaller or equal to the number of 1's in the left node then go left in the segment tree,

otherwise go to the right child and decrease A with the number of 1's in the left child; o(nlogn)

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9 years ago, hide # |
 
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Can you provide a link for that problem?

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9 years ago, hide # |
 
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Answer queries in the reverse order

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9 years ago, hide # |
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Can be done in O(logN) per query using a BIT.

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    9 years ago, hide # ^ |
     
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    how is it O(logN) ? , considering you are doing binary search on bit .isn't it O(log^2N)

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      9 years ago, hide # ^ |
       
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      You can binary search on the bit itself. Check topcoder tutorial for BIT.

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