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.
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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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)
Can you provide a link for that problem?
Answer queries in the reverse order
Can be done in O(logN) per query using a BIT.
how is it O(logN) ? , considering you are doing binary search on bit .isn't it O(log^2N)
You can binary search on the bit itself. Check topcoder tutorial for BIT.