Queries for number of distinct elements in a subarray

Difficulty Level : Hard
Last Updated : 23 Jan, 2020

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)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use Binary Indexed Tree

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.
Step 2 : Sort all the queries in ascending order of their right end r.
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]].
Step 4 : Set last_visit[a[i]] = i and update the bit array bit array with value 1 at idx i.
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.

C++

// 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;
}

Java

// Java code to find number of distinct numbers 
// in a subarray 
import java.io.*;
import java.util.*;
  
class GFG 
{
    static int MAX = 1000001;
  
    // structure to store queries
    static class Query 
    {
        int l, r, idx;
    }
  
    // updating the bit array
    static void update(int idx, int val, 
                        int bit[], int n) 
    {
        for (; idx <= n; idx += idx & -idx)
            bit[idx] += val;
    }
  
    // querying the bit array
    static int query(int idx, int bit[], int n) 
    {
        int sum = 0;
        for (; idx > 0; idx -= idx & -idx)
            sum += bit[idx];
        return sum;
    }
  
    static void answeringQueries(int[] arr, int n, 
                                Query[] queries, int q)
    {
  
        // initialising bit array
        int[] bit = new int[n + 1];
        Arrays.fill(bit, 0);
  
        // holds the rightmost index of any number
        // as numbers of a[i] are less than or equal to 10^6
        int[] last_visit = new int[MAX];
        Arrays.fill(last_visit, -1);
  
        // answer for each query
        int[] ans = new int[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++)
            System.out.println(ans[i]);
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        int a[] = { 1, 1, 2, 1, 3 };
        int n = a.length;
        Query[] queries = new Query[3];
        for (int i = 0; i < 3; i++)
            queries[i] = new Query();
        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 = queries.length;
        Arrays.sort(queries, new Comparator<Query>() 
        {
            public int compare(Query x, Query y)
            {
                if (x.r < y.r)
                    return -1;
                else if (x.r == y.r)
                    return 0;
                else
                    return 1;
            }
        });
        answeringQueries(a, n, queries, q);
    }
}
  
// This code is contributed by
// sanjeev2552

Python3

# Python3 code to find number of 
# distinct numbers in a subarray
MAX = 1000001
  
# structure to store queries
class Query:
    def __init__(self, l, r, idx):
        self.l = l
        self.r = r
        self.idx = idx
  
# updating the bit array
def update(idx, val, bit, n):
    while idx <= n:
        bit[idx] += val
        idx += idx & -idx
  
# querying the bit array
def query(idx, bit, n):
    summ = 0
    while idx:
        summ += bit[idx]
        idx -= idx & -idx
    return summ
  
def answeringQueries(arr, n, queries, q):
  
    # initialising bit array
    bit = [0] * (n + 1)
  
    # holds the rightmost index of 
    # any number as numbers of a[i]
    # are less than or equal to 10^6
    last_visit = [-1] * MAX
  
    # answer for each query
    ans = [0] * q
  
    query_counter = 0
    for i in range(n):
  
        # 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 and 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 += 1
  
    # print answer for each query
    for i in range(q):
        print(ans[i])
  
# Driver Code
if __name__ == "__main__":
    a = [1, 1, 2, 1, 3]
    n = len(a)
    queries = [Query(0, 4, 0), 
               Query(1, 3, 1), 
               Query(2, 4, 2)]
    q = len(queries)
  
    queries.sort(key = lambda x: x.r)
    answeringQueries(a, n, queries, q)
  
# This code is contributed by
# sanjeev2552

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes