Distinct elements in subarray using Mo’s Algorithm

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.
  • Sort the queries in the same order as they were provided earlier and print their stored results
  • Adding elements()

    Removing elements()

    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 :

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    // 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 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 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;
    }
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    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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