Given an 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] <=

Examples :

Input : a[] = {1, 1, 2, 1, 2, 3} q = 3 0 4 1 3 2 5 Output : 2 2 3 In query 1, number of distinct integers in a[0...4] is 2 (1, 2) 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..5] is 3 (1, 2, 3) Input : a[] = {7, 3, 5, 9, 7, 6, 4, 3, 2} q = 4 1 5 0 4 0 7 1 8 output : 5 4 6 7

Let a[0…n-1] be input array and q[0..m-1] be array of queries.

**Approach :**

- Sort all queries in a way that queries with L values from 0 to are put together, then all queries from to , and so on. All queries within a block are sorted in increasing order of R values.
- Initialize an array freq[] of size with 0 . freq[] array keep count of frequencies of all the elements in lying in a given range.
- Process all queries one by one in a way that every query uses number of different elements and frequency array computed in previous query and stores the result in structure.
- Let ‘curr_Diff_element’ be number of different elements of previous query.
- Remove extra elements of previous query. For example if previous query is [0, 8] and current query is [3, 9], then remove a[0], a[1] and a[2]
- Add new elements of current query. In the same example as above, add a[9].
- Sort the queries in the same order as they were provided earlier and print their stored results

**Adding elements()**

- Increase the frequency of element to be added(freq[a[i]]) by 1.
- If frequency of element a[i] is 1.Increase curr_diff_element by 1 as 1 new element has been added in range.

**Removing elements() **

- Decrease frequency of element to be removed (a[i]) by 1.
- if frequency of an element a[i] is 0.Just decrease curr_diff_element by 1 as 1 element has been completely removed from the range.

**Note : ** In this algorithm, in step 2, index variable for R change at most O(n * ) times throughout the run and same for L changes its value at most O(m * ) times. All these bounds are possible only because sorted queries first in blocks of size.

The preprocessing part takes O(m Log m) time.

Processing all queries takes O(n * ) + O(m * ) = O((m+n) *) time.

**Below is the implementation of above approach :**

// Program to compute no. of different elements // of ranges for different range queries #include <bits/stdc++.h> using namespace std; // Used in frequency array (maximum value of an // array element). const int MAX = 1000000; // Variable to represent block size. This is made // global so compare() of sort can use it. int block; // Structure to represent a query range and to store // index and result of a particular query range struct Query { int L, R, index, result; }; // Function used to sort all queries so that all queries // of same block are arranged together and within a block, // queries are sorted in increasing order of R values. bool compare(Query x, Query y) { // Different blocks, sort by block. if (x.L / block != y.L / block) return x.L / block < y.L / block; // Same block, sort by R value return x.R < y.R; } // Function used to sort all queries in order of their // index value so that results of queries can be printed // in same order as of input bool compare1(Query x, Query y) { return x.index < y.index; } // calculate distinct elements of all query ranges. // m is number of queries n is size of array a[]. void queryResults(int a[], int n, Query q[], int m) { // Find block size block = (int)sqrt(n); // Sort all queries so that queries of same // blocks are arranged together. sort(q, q + m, compare); // Initialize current L, current R and current // different elements int currL = 0, currR = 0; int curr_Diff_elements = 0; // Initialize frequency array with 0 int freq[MAX] = { 0 }; // Traverse through all queries for (int i = 0; i < m; i++) { // L and R values of current range int L = q[i].L, R = q[i].R; // Remove extra elements of previous range. // For example if previous range is [0, 3] // and current range is [2, 5], then a[0] // and a[1] are subtracted while (currL < L) { // element a[currL] is removed freq[a[currL]]--; if (freq[a[currL]] == 0) curr_Diff_elements--; currL++; } // Add Elements of current Range // Note:- during addition of the left // side elements we have to add currL-1 // because currL is already in range while (currL > L) { freq[a[currL - 1]]++; // include a element if it occurs first time if (freq[a[currL - 1]] == 1) curr_Diff_elements++; currL--; } while (currR <= R) { freq[a[currR]]++; // include a element if it occurs first time if (freq[a[currR]] == 1) curr_Diff_elements++; currR++; } // Remove elements of previous range. For example // when previous range is [0, 10] and current range // is [3, 8], then a[9] and a[10] are subtracted // Note:- Basically for a previous query L to R // currL is L and currR is R+1. So during removal // of currR remove currR-1 because currR was // never included while (currR > R + 1) { // element a[currL] is removed freq[a[currR - 1]]--; // if ocurrence of a number is reduced // to zero remove it from list of // different elements if (freq[a[currR - 1]] == 0) curr_Diff_elements--; currR--; } q[i].result = curr_Diff_elements; } } // print the result of all range queries in // initial order of queries void printResults(Query q[], int m) { sort(q, q + m, compare1); for (int i = 0; i < m; i++) { cout << "Number of different elements" << " in range " << q[i].L << " to " << q[i].R << " are " << q[i].result << endl; } } // Driver program int main() { int a[] = { 1, 1, 2, 1, 3, 4, 5, 2, 8 }; int n = sizeof(a) / sizeof(a[0]); Query q[] = { { 0, 4, 0, 0 }, { 1, 3, 1, 0 }, { 2, 4, 2, 0 } }; int m = sizeof(q) / sizeof(q[0]); queryResults(a, n, q, m); printResults(q, m); return 0; }

**Output:**

Number of different elements in range 0 to 4 are 3 Number of different elements in range 1 to 3 are 2 Number of different elements in range 2 to 4 are 3

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