Queries for number of distinct elements in a subarray

Last Updated : 27 Apr, 2023

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

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

using System;
using System.Linq;

class GFG {
    static int MAX = 1000001;
    // structure to store queries
    public class Query {
        public int l, r, idx;
    }

    // updating the bit array
    public static void update(int idx, int val, int[] bit,
                              int n)
    {
        for (; idx <= n; idx += idx & -idx)
            bit[idx] += val;
    }

    // querying the bit array
    public static int query(int idx, int[] bit, int n)
    {
        int sum = 0;
        for (; idx > 0; idx -= idx & -idx)
            sum += bit[idx];
        return sum;
    }

    public static void answeringQueries(int[] arr, int n,
                                        Query[] queries,
                                        int q)
    {
        // initialising bit array
        int[] bit = new int[n + 1];
        Array.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];
        Array.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++)
            Console.WriteLine(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;
        Array.Sort(queries, (x, y) => x.r - y.r);
        answeringQueries(a, n, queries, q);
    }
}

JavaScript

<script>

// JavaScript code to find number of
// distinct numbers in a subarray
const MAX = 1000001

// structure to store queries
class Query{
    constructor(l, r, idx){
        this.l = l
        this.r = r
        this.idx = idx
    }
}

// updating the bit array
function update(idx, val, bit, n){
    while(idx <= n){
        bit[idx] += val
        idx += idx & -idx
    }
}

// querying the bit array
function query(idx, bit, n){
    let summ = 0
    while(idx){
        summ += bit[idx]
        idx -= idx & -idx
    }
    return summ
}

function answeringQueries(arr, n, queries, q){

    // initialising bit array
    let bit = new Array(n+1).fill(0)

    // holds the rightmost index of
    // any number as numbers of a[i]
    // are less than or equal to 10^6
    let last_visit = new Array(MAX).fill(-1)

    // answer for each query
    let ans = new Array(q).fill(0);

    let query_counter = 0
    for(let 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 += 1
        }
    }

    // print answer for each query
    for(let i=0;i<q;i++)
        document.write(ans[i],"</br>")
}

// Driver Code

let a = [1, 1, 2, 1, 3]
let n = a.length
let queries = [new Query(0, 4, 0),new Query(1, 3, 1),new Query(2, 4, 2)]
let q = queries.length

queries.sort((x,y) => x.r-y.r)
answeringQueries(a, n, queries, q)

// This code is contributed by shinjanpatra

</script>

Output:

3
2
3

Time Complexity: O((n+q)*log(n))

Auxiliary Space: O(n+q+MAX)

Queries for number of distinct elements in a subarray

Ayush Jha

Improve

Article Tags :

Practice Tags :

Advanced Data Structure

Queries for number of distinct elements in a subarray

Similar Reads

Thank You!

What kind of Experience do you want to share?