Algorithm
 Initialize an array ‘prev’ of size 10^6 with 1 and ‘BIT’ of size ‘N’+1 with 0.
 Sort the queries in increasing order of their right end ‘r’.
 Traverse the array and check if ‘prev[arr[i]]’ is 1 or not.
 If it is 1, update the value of BIT at index ‘i’ with 1.
 Else, update the value of BIT at index ‘prev[arr[i]]’ with 0 and at index ‘i’ with 1.
 Answer the queries whose right end ‘r’ equals, by performing two queries on BIT  query(‘r’)  query(‘l’1).
C++ Code
#include<bits/stdc++.h>
using namespace std;
const int MAX = 10001;
// Comparator to sort queries
bool cmp(vector<int> x, vector<int> y){
return x[1] < y[1];
}
// Update function of BIT
void update(int i, int val, int bit[], int n){
for (; i <= n; i += i&i){
bit[i] += val;
}
}
// Query function of BIT
int query(int i, int bit[], int n){
int sum = 0;
for (; i>0; i=i&i)
sum += bit[i];
return sum;
}
void solve(int arr[], int n, vector<int> queries[], int q)
{
sort(queries, queries+q, cmp);
// Initialising BIT
int bit[n+1];
memset(bit, 0, sizeof(bit));
// Initializing prev
int prev[MAX];
memset(prev, 1, sizeof(prev));
// Array to store answer for all queries
int ans[q];
memset(ans, 1, sizeof(ans));
int qind = 0;
// Iterating over input array
for (int i=0; i<n; i++){
// If prev is not 1, updating BIT
if (prev[arr[i]] !=1){
update (prev[arr[i]] + 1, 1, bit, n);
}
// Setting prev[arr[i]] as i, updating BIT
prev[arr[i]] = i;
update(i + 1, 1, bit, n);
// Checking if r of any query is equal to i
while (qind < q && queries[qind][1] == i+1){
int l = queries[qind][0], r = queries[qind][1];
int idx = queries[qind][2];
// Calculating the number of distinct elements in the subarray
ans[idx] = query(r, bit, n) query(l1, bit, n);
qind++;
}
}
// Printing the number of distinct elements in the subarray
for (int i=0; i<q; i++){
cout << ans[i] << ' ';
}
cout<<endl;
}
// Driver code
int main()
{
int n = 5;
int a[] = {1, 2, 1, 3, 2};
int q = 3;
// queries of the form  {l, r, idx}
vector<int>queries[] = { {1, 3, 0}, {4, 5, 1}, {2, 5, 2} };
solve(a, n, queries, q);
return 0;
}
Output:
2 2 3
Algorithm Complexity
Time Complexity
We have firstly sorted the queries which will take O(QlogQ) time.
Then, for every element, we are performing an update on BIT which will take O(NlogN) time. Also, for each query element, we call the query() function of BIT two times to find the number of distinct elements in the subarray. This will take O(QlogN) time.
Therefore, total time complexity will be O(QlogQ + (Q + N)logN)
Space Complexity
We are creating a BIT array of size N+1. Also, we are creating a ‘prev’ array to store the previous index of an element. This will also have N+1 size as arr[i]<=N.
Therefore, the total space complexity of the above approach is O(N).
Check out this problem  Subarray With 0 Sum
FAQs

What is a Fenwick Tree?
A Binary Indexed Tree or Fenwick tree is a data structure that can efficiently update elements and calculate prefix sums of a list of numbers. It allows both the operations to be performed in time complexity of O(logN).

What is an offline query?
An offline query means that we can first store all the queries and then solve them in any order to calculate the answer more efficiently. This can be used only when the answer for one query is independent of the other.

Why do we decide to use offline queries in this problem?
In this problem, we had to find the number of distinct elements in the subarray between L and R for multiple queries. Since we have to count distinct elements, we have to consider each element only once. Now suppose for a single query ‘l’ and ‘r’, we need to consider only the elements at index smaller than ‘r’. Also, for each such element, we just have to check if its rightmost occurrence is at an index greater than ‘l’ or not. Therefore, we decided to use the offline queries approach.
Key Takeaways
This article discussed finding the number of distinct elements in the subarray from ‘l’ to ‘r’ for multiple queries. We also discussed the time and space complexity of our solution. If you want to solve more such problems, you can visit Coding Ninjas Studio.
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Until then, all the best for your future endeavors, and Keep Coding.