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] <= 106Examples:
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 |
C#
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)