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Queries for number of distinct elements in a subarray
• Difficulty Level : Hard
• Last Updated : 23 Jan, 2020

Given a array ‘a[]’ of size n and number of queries q. Each query can be represented by two integers l and r. Your task is to print the number of distinct integers in the subarray l to r.
Given a[i] <= 106
Examples:

```Input : a[] = {1, 1, 2, 1, 3}
q = 3
0 4
1 3
2 4
Output :3
2
3
In query 1, number of distinct integers
in a[0...4] is 3 (1, 2, 3)
In query 2, number of distinct integers
in a[1..3] is 2 (1, 2)
In query 3, number of distinct integers
in a[2..4] is 3 (1, 2, 3)
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use Binary Indexed Tree

1. Step 1 : Take an array last_visit of size 10^6 where last_visit[i] holds the rightmost index of the number i in the array a. Initialize this array as -1.
2. Step 2 : Sort all the queries in ascending order of their right end r.
3. Step 3 : Create a Binary Indexed Tree in an array bit[]. Start traversing the array ‘a’ and queries simultaneously and check if last_visit[a[i]] is -1 or not. If it is not, update the bit array with value -1 at the idx last_visit[a[i]].
4. Step 4 : Set last_visit[a[i]] = i and update the bit array bit array with value 1 at idx i.
5. Step 5 : Answer all the queries whose value of r is equal to i by querying the bit array. This can be easily done as queries are sorted.

## C++

 `// C++ code to find number of distinct numbers``// in a subarray``#include``using` `namespace` `std;`` ` `const` `int` `MAX = 1000001;`` ` `// structure to store queries``struct` `Query``{``    ``int` `l, r, idx;``};`` ` `// cmp function to sort queries according to r``bool` `cmp(Query x, Query y)``{``    ``return` `x.r < y.r;``}`` ` `// updating the bit array``void` `update(``int` `idx, ``int` `val, ``int` `bit[], ``int` `n)``{``    ``for` `(; idx <= n; idx += idx&-idx)``        ``bit[idx] += val;``}`` ` `// querying the bit array``int` `query(``int` `idx, ``int` `bit[], ``int` `n)``{``    ``int` `sum = 0;``    ``for` `(; idx>0; idx-=idx&-idx)``        ``sum += bit[idx];``    ``return` `sum;``}`` ` `void` `answeringQueries(``int` `arr[], ``int` `n, Query queries[], ``int` `q)``{``    ``// initialising bit array``    ``int` `bit[n+1];``    ``memset``(bit, 0, ``sizeof``(bit));`` ` `    ``// holds the rightmost index of any number``    ``// as numbers of a[i] are less than or equal to 10^6``    ``int` `last_visit[MAX];``    ``memset``(last_visit, -1, ``sizeof``(last_visit));`` ` `    ``// answer for each query``    ``int` `ans[q];``    ``int` `query_counter = 0;``    ``for` `(``int` `i=0; i

## Java

 `// Java code to find number of distinct numbers ``// in a subarray ``import` `java.io.*;``import` `java.util.*;`` ` `class` `GFG ``{``    ``static` `int` `MAX = ``1000001``;`` ` `    ``// structure to store queries``    ``static` `class` `Query ``    ``{``        ``int` `l, r, idx;``    ``}`` ` `    ``// updating the bit array``    ``static` `void` `update(``int` `idx, ``int` `val, ``                        ``int` `bit[], ``int` `n) ``    ``{``        ``for` `(; idx <= n; idx += idx & -idx)``            ``bit[idx] += val;``    ``}`` ` `    ``// querying the bit array``    ``static` `int` `query(``int` `idx, ``int` `bit[], ``int` `n) ``    ``{``        ``int` `sum = ``0``;``        ``for` `(; idx > ``0``; idx -= idx & -idx)``            ``sum += bit[idx];``        ``return` `sum;``    ``}`` ` `    ``static` `void` `answeringQueries(``int``[] arr, ``int` `n, ``                                ``Query[] queries, ``int` `q)``    ``{`` ` `        ``// initialising bit array``        ``int``[] bit = ``new` `int``[n + ``1``];``        ``Arrays.fill(bit, ``0``);`` ` `        ``// holds the rightmost index of any number``        ``// as numbers of a[i] are less than or equal to 10^6``        ``int``[] last_visit = ``new` `int``[MAX];``        ``Arrays.fill(last_visit, -``1``);`` ` `        ``// answer for each query``        ``int``[] ans = ``new` `int``[q];``        ``int` `query_counter = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) ``        ``{`` ` `            ``// If last visit is not -1 update -1 at the``            ``// idx equal to last_visit[arr[i]]``            ``if` `(last_visit[arr[i]] != -``1``)``                ``update(last_visit[arr[i]] + ``1``, -``1``, bit, n);`` ` `            ``// Setting last_visit[arr[i]] as i and updating``            ``// the bit array accordingly``            ``last_visit[arr[i]] = i;``            ``update(i + ``1``, ``1``, bit, n);`` ` `            ``// If i is equal to r of any query store answer``            ``// for that query in ans[]``            ``while` `(query_counter < q && queries[query_counter].r == i) ``            ``{``                ``ans[queries[query_counter].idx] = ``                        ``query(queries[query_counter].r + ``1``, bit, n)``                        ``- query(queries[query_counter].l, bit, n);``                ``query_counter++;``            ``}``        ``}`` ` `        ``// print answer for each query``        ``for` `(``int` `i = ``0``; i < q; i++)``            ``System.out.println(ans[i]);``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args) ``    ``{``        ``int` `a[] = { ``1``, ``1``, ``2``, ``1``, ``3` `};``        ``int` `n = a.length;``        ``Query[] queries = ``new` `Query[``3``];``        ``for` `(``int` `i = ``0``; i < ``3``; i++)``            ``queries[i] = ``new` `Query();``        ``queries[``0``].l = ``0``;``        ``queries[``0``].r = ``4``;``        ``queries[``0``].idx = ``0``;``        ``queries[``1``].l = ``1``;``        ``queries[``1``].r = ``3``;``        ``queries[``1``].idx = ``1``;``        ``queries[``2``].l = ``2``;``        ``queries[``2``].r = ``4``;``        ``queries[``2``].idx = ``2``;``        ``int` `q = queries.length;``        ``Arrays.sort(queries, ``new` `Comparator() ``        ``{``            ``public` `int` `compare(Query x, Query y)``            ``{``                ``if` `(x.r < y.r)``                    ``return` `-``1``;``                ``else` `if` `(x.r == y.r)``                    ``return` `0``;``                ``else``                    ``return` `1``;``            ``}``        ``});``        ``answeringQueries(a, n, queries, q);``    ``}``}`` ` `// This code is contributed by``// sanjeev2552`

## Python3

 `# Python3 code to find number of ``# distinct numbers in a subarray``MAX` `=` `1000001`` ` `# structure to store queries``class` `Query:``    ``def` `__init__(``self``, l, r, idx):``        ``self``.l ``=` `l``        ``self``.r ``=` `r``        ``self``.idx ``=` `idx`` ` `# updating the bit array``def` `update(idx, val, bit, n):``    ``while` `idx <``=` `n:``        ``bit[idx] ``+``=` `val``        ``idx ``+``=` `idx & ``-``idx`` ` `# querying the bit array``def` `query(idx, bit, n):``    ``summ ``=` `0``    ``while` `idx:``        ``summ ``+``=` `bit[idx]``        ``idx ``-``=` `idx & ``-``idx``    ``return` `summ`` ` `def` `answeringQueries(arr, n, queries, q):`` ` `    ``# initialising bit array``    ``bit ``=` `[``0``] ``*` `(n ``+` `1``)`` ` `    ``# holds the rightmost index of ``    ``# any number as numbers of a[i]``    ``# are less than or equal to 10^6``    ``last_visit ``=` `[``-``1``] ``*` `MAX`` ` `    ``# answer for each query``    ``ans ``=` `[``0``] ``*` `q`` ` `    ``query_counter ``=` `0``    ``for` `i ``in` `range``(n):`` ` `        ``# If last visit is not -1 update -1 at the``        ``# idx equal to last_visit[arr[i]]``        ``if` `last_visit[arr[i]] !``=` `-``1``:``            ``update(last_visit[arr[i]] ``+` `1``, ``-``1``, bit, n)`` ` `        ``# Setting last_visit[arr[i]] as i and ``        ``# updating the bit array accordingly``        ``last_visit[arr[i]] ``=` `i``        ``update(i ``+` `1``, ``1``, bit, n)`` ` `        ``# If i is equal to r of any query store answer``        ``# for that query in ans[]``        ``while` `query_counter < q ``and` `queries[query_counter].r ``=``=` `i:``            ``ans[queries[query_counter].idx] ``=` `\``                        ``query(queries[query_counter].r ``+` `1``, bit, n) ``-` `\``                        ``query(queries[query_counter].l, bit, n)``            ``query_counter ``+``=` `1`` ` `    ``# print answer for each query``    ``for` `i ``in` `range``(q):``        ``print``(ans[i])`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``a ``=` `[``1``, ``1``, ``2``, ``1``, ``3``]``    ``n ``=` `len``(a)``    ``queries ``=` `[Query(``0``, ``4``, ``0``), ``               ``Query(``1``, ``3``, ``1``), ``               ``Query(``2``, ``4``, ``2``)]``    ``q ``=` `len``(queries)`` ` `    ``queries.sort(key ``=` `lambda` `x: x.r)``    ``answeringQueries(a, n, queries, q)`` ` `# This code is contributed by``# sanjeev2552`
Output:
```3
2
3
```

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