Queries for number of distinct elements in a subarray

Given a 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] <= 106
Examples:

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

The idea is to use Binary Indexed Tree

  1. Step 1 : Take an array last_visit of size 10^6 where last_visit[i] holds the rightmost index of the number i in the array a. Initialize this array as -1.
  2. Step 2 : Sort all the queries in ascending order of their right end r.
  3. Step 3 : Create a Binary Indexed Tree in an array bit[]. Start traversing the array ‘a’ and queries simultaneously and check if last_visit[a[i]] is -1 or not. If it is not, update the bit array with value -1 at the idx last_visit[a[i]].
  4. Step 4 : Set last_visit[a[i]] = i and update the bit array bit array with value 1 at idx i.
  5. Step 5 : Answer all the queries whose value of r is equal to i by querying the bit array. This can be easily done as queries are sorted.
filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ code to find number of distinct numbers
// in a subarray
#include<bits/stdc++.h>
using namespace std;
  
const int MAX = 1000001;
  
// structure to store queries
struct Query
{
    int l, r, idx;
};
  
// cmp function to sort queries according to r
bool cmp(Query x, Query y)
{
    return x.r < y.r;
}
  
// updating the bit array
void update(int idx, int val, int bit[], int n)
{
    for (; idx <= n; idx += idx&-idx)
        bit[idx] += val;
}
  
// querying the bit array
int query(int idx, int bit[], int n)
{
    int sum = 0;
    for (; idx>0; idx-=idx&-idx)
        sum += bit[idx];
    return sum;
}
  
void answeringQueries(int arr[], int n, Query queries[], int q)
{
    // initialising bit array
    int bit[n+1];
    memset(bit, 0, sizeof(bit));
  
    // holds the rightmost index of any number
    // as numbers of a[i] are less than or equal to 10^6
    int last_visit[MAX];
    memset(last_visit, -1, sizeof(last_visit));
  
    // answer for each query
    int ans[q];
    int query_counter = 0;
    for (int i=0; i<n; i++)
    {
        // If last visit is not -1 update -1 at the
        // idx equal to last_visit[arr[i]]
        if (last_visit[arr[i]] !=-1)
            update (last_visit[arr[i]] + 1, -1, bit, n);
  
        // Setting last_visit[arr[i]] as i and updating
        // the bit array accordingly
        last_visit[arr[i]] = i;
        update(i + 1, 1, bit, n);
  
        // If i is equal to r of any query  store answer
        // for that query in ans[]
        while (query_counter < q && queries[query_counter].r == i)
        {
            ans[queries[query_counter].idx] =
                     query(queries[query_counter].r + 1, bit, n)-
                     query(queries[query_counter].l, bit, n);
            query_counter++;
        }
    }
  
    // print answer for each query
    for (int i=0; i<q; i++)
        cout << ans[i] << endl;
}
  
// driver code
int main()
{
    int a[] = {1, 1, 2, 1, 3};
    int n = sizeof(a)/sizeof(a[0]);
    Query queries[3];
    queries[0].l = 0;
    queries[0].r = 4;
    queries[0].idx = 0;
    queries[1].l = 1;
    queries[1].r = 3;
    queries[1].idx = 1;
    queries[2].l = 2;
    queries[2].r = 4;
    queries[2].idx = 2;
    int q = sizeof(queries)/sizeof(queries[0]);
    sort(queries, queries+q, cmp);
    answeringQueries(a, n, queries, q);
    return 0;
}

chevron_right






                        filter_none








Output:


3
2
3



This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts: