Is it possible to query number of distinct integers in a range in O(lg N)?

Question

I have read through some tutorials about two common data structure which can achieve range update and query in O(lg N): Segment tree and Binary Indexed Tree (BIT / Fenwick T

Answer 1

There is a well-known offline method to solve this problem. If you have n size array and q queries on it and in each query, you need to know the count of distinct number in that range then you can solve this whole thing in O(n log n + q log n) time complexity. Which is similar to solve every query in O(log n) time.

Let's solve the problem using the RSQ( Range sum query) technique. For the RSQ technique, you can use a segment tree or BIT. Let's discuss the segment tree technique.

For solving this problem you need an offline technique and a segment tree. Now, what is an offline technique?? The offline technique is doing something offline. In problem-solving an example of the offline technique is, You take input all queries first and then reorder them is a way so that you can answer them correctly and easily and finally output the answers in the given input order.

Solution Idea:

First, take input for a test case and store the given n numbers in an array. Let the array name is array[] and take input q queries and store them in a vector v. where every element of v hold three field- l, r, idx. where l is the start point of a query and r is the endpoint of a query and idx is the number of queries. like this one is n^th query. Now sort the vector v on the basis of the endpoint of a query. Let we have a segment tree which can store the information of at least 10^5 element. and we also have an areay called last[100005]. which stores the last position of a number in the array[].

Initially, all elements of the tree are zero and all elements of the last are -1. now run a loop on the array[]. now inside the loop, you have to check this thing for every index of array[].

last[array[i]] is -1 or not? if it is -1 then write last[array[i]]=i and call update() function of which will add +1 in the last[array[i]] th position of segment tree. if last[array[i]] is not -1 then call update() function of segment tree which will subtract 1 or add -1 in the last[array[i]] th position of segment tree. Now you need to store current position as last position for future. so that you need to write last[array[i]]=i and call update() function which will add +1 in the last[array[i]] th position of segment tree.

Now you have to check whether a query is finished in the current index. that is if(v[current].r==i). if this is true then call query() function of segment tree which will return and sum of the range v[current].l to v[current].r and store the result in the v[current].idx^th index of the answer[] array. you also need to increment the value of current by 1. 6. Now print the answer[] array which contains your final answer in the given input order.

the complexity of the algorithm is O(n log n).