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Distinct elements in subarray using Mo’s Algorithm
• Difficulty Level : Hard
• Last Updated : 11 Jul, 2019

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 :

1. 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.
2. Initialize an array freq[] of size with 0 . freq[] array keep count of frequencies of all the elements in lying in a given range.
3. 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, a and a
• Add new elements of current query. In the same example as above, add a.
4. Sort the queries in the same order as they were provided earlier and print their stored results

• 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 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 rangestruct 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 inputbool 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         // and a 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 and a 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 occurrence 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 queriesvoid 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 programint main(){    int a[] = { 1, 1, 2, 1, 3, 4, 5, 2, 8 };    int n = sizeof(a) / sizeof(a);    Query q[] = { { 0, 4, 0, 0 }, { 1, 3, 1, 0 },                  { 2, 4, 2, 0 } };    int m = sizeof(q) / sizeof(q);    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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