// Program to compute no. of different elements 
// of ranges for different range queries 
#include <bits/stdc++.h> 
using namespace std; 
#define ll long long 
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
// 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<<q[i].result << endl; 
    } 
} 

// Driver program 
int main() 
{ 
   // int a[] = { 1, 1, 2, 1, 3, 4, 5, 2, 8 }; 
    //i//nt n = sizeof(a) / sizeof(a[0]);
     fast;
     int n,q;
    cin>>n>>q;
     int a[n];
     int i;
    for(i=0;i<n;i++)
        cin>>a[i];
    Query qq[q];
     int x,y;
    for(i=0;i<q;i++)
    {
        cin>>x>>y;
        qq[i].L=x-1;qq[i].R=y-1;qq[i].index=i;qq[i].result=0;
    }
    //int m = sizeof(q) / sizeof(q[0]); 
    queryResults(a, n, qq, q); 
    printResults(qq, q); 
    return 0; 
} 


Measured memory_bytes: 17348