Distinct elements in subarray using Mo's Algorithm - GeeksforGeeks

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 :
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
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].
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 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[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 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[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;}
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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