Queries for number of distinct integers in Suffix

Given an array of N elements and Q queries, where each query contains an integer K. For each query, the task is to find the number of distinct integers in the suffix from K^th element to N^th element.

Examples:

Input :
N=5, Q=3
arr[] = {2, 4, 6, 10, 2}
1
3
2
Output :
4
3
4

Approach:
The problem can be solved by precomputing the solution for all index from N-1 to 0. The idea is to use an unordered_set in C++ as set does not allow duplicate elements.
Traverse the array from end and add the current element into a set and the answer for the current index would be the size of the set. So, store the size of set at current index in an auxiliary array say suffixCount. 

Below is the implementation of the above approach: 

C++

// C++ program to find distinct
// elements in suffix

#include <bits/stdc++.h>
using namespace std;

// Function to answer queries for distinct
// count in Suffix
void printQueries(int n, int a[], int q, int qry[])
{
    // Set to keep the distinct elements
    unordered_set<int> occ;

    int suffixCount[n + 1];

    // Precompute the answer for each suffix
    for (int i = n - 1; i >= 0; i--) {
        occ.insert(a[i]);
        suffixCount[i + 1] = occ.size();
    }

    // Print the precomputed answers
    for (int i = 0; i < q; i++)
        cout << suffixCount[qry[i]] << endl;
}

// Driver Code
int main()
{
    int n = 5, q = 3;
    int a[n] = { 2, 4, 6, 10, 2 };
    int qry[q] = { 1, 3, 2 };

    printQueries(n, a, q, qry);

    return 0;
}

Java

// Java program to find distinct
// elements in suffix
import java.util.*;

class GFG
{


// Function to answer queries for distinct
// count in Suffix
static void printQueries(int n, int a[], int q, int qry[])
{
    // Set to keep the distinct elements
    HashSet<Integer> occ = new HashSet<>();

    int []suffixCount = new int[n + 1];

    // Precompute the answer for each suffix
    for (int i = n - 1; i >= 0; i--)
    {
        occ.add(a[i]);
        suffixCount[i + 1] = occ.size();
    }

    // Print the precomputed answers
    for (int i = 0; i < q; i++)
        System.out.println(suffixCount[qry[i]]);
}

// Driver Code
public static void main(String args[])
{
    int n = 5, q = 3;
    int a[] = { 2, 4, 6, 10, 2 };
    int qry[] = { 1, 3, 2 };

    printQueries(n, a, q, qry);
}
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to find distinct
# elements in suffix

# Function to answer queries for distinct
# count in Suffix
def printQueries(n, a, q, qry):

    # Set to keep the distinct elements
    occ = dict()

    suffixCount = [0 for i in range(n + 1)]

    # Precompute the answer for each suffix
    for i in range(n - 1, -1, -1):
        occ[a[i]] = 1
        suffixCount[i + 1] = len(occ)

    # Print the precomputed answers
    for i in range(q):
        print(suffixCount[qry[i]])

# Driver Code
n = 5
q = 3
a = [2, 4, 6, 10, 2]
qry = [1, 3, 2]

printQueries(n, a, q, qry)

# This code is contributed by Mohit Kumar 29

C#

// C# program to find distinct
// elements in suffix
using System;
using System.Collections.Generic;

public class GFG
{


// Function to answer queries for distinct
// count in Suffix
static void printQueries(int n, int []a, int q, int []qry)
{
    // Set to keep the distinct elements
    HashSet<int> occ = new HashSet<int>();

    int []suffixCount = new int[n + 1];

    // Precompute the answer for each suffix
    for (int i = n - 1; i >= 0; i--)
    {
        occ.Add(a[i]);
        suffixCount[i + 1] = occ.Count;
    }

    // Print the precomputed answers
    for (int i = 0; i < q; i++)
        Console.WriteLine(suffixCount[qry[i]]);
}

// Driver Code
public static void Main(String []args)
{
    int n = 5, q = 3;
    int []a = { 2, 4, 6, 10, 2 };
    int []qry = { 1, 3, 2 };

    printQueries(n, a, q, qry);
}
}

// This code is contributed by Princi Singh

PHP

<?php
// PHP program to find distinct
// elements in suffix

// Function to answer queries for distinct
// count in Suffix
function printQueries($n, $a, $q, $qry)
{
    // Set to keep the distinct elements
    $occ = array();

    $suffixCount = array_fill(0, $n + 1, 0);

    // Precompute the answer for each suffix
    for ($i = $n - 1; $i >= 0; $i--)
    {
        array_push($occ, $a[$i]);

        # give array of distinct element
        $occ = array_unique($occ);

        $suffixCount[$i + 1] = sizeof($occ);
    }


    // Print the precomputed answers
    for ($i = 0; $i < $q; $i++)
        echo $suffixCount[$qry[$i]], "\n";
}

// Driver Code
$n = 5;
$q = 3;
$a = array(2, 4, 6, 10, 2);
$qry = array(1, 3, 2);

printQueries($n, $a, $q, $qry);

// This code is contributed by Ryuga
?>

Javascript

<script>

// JavaScript program to find distinct
// elements in suffix

// Function to answer queries for distinct
// count in Suffix
function printQueries(n,a,q,qry)
{
    // Set to keep the distinct elements
    let occ = new Set();

    let suffixCount = new Array(n + 1);

    // Precompute the answer for each suffix
    for (let i = n - 1; i >= 0; i--)
    {
        occ.add(a[i]);
        suffixCount[i + 1] = occ.size;
    }

    // Print the precomputed answers
    for (let i = 0; i < q; i++)
        document.write(suffixCount[qry[i]]+"<br>");
}

// Driver Code
let  n = 5, q = 3;
let a=[2, 4, 6, 10, 2];
let qry=[ 1, 3, 2 ];
printQueries(n, a, q, qry);


// This code is contributed by patel2127

</script>

Output:

4
3
4

Time Complexity: O(N) 
Auxiliary Space: O(N)
Approach without STL: Since queries need to be addressed, precomputation needs to be done. Otherwise, a simple traversal over the suffix would have sufficed.
An auxiliary array aux[] is maintained from the right side of the array. An array mark[] stores whether an element has occurred or not in the suffix traversed yet. If the element has not occurred, then update aux[i] = aux[i+1] + 1. Otherwise aux[i] = aux[i+1].
Answer for every query would be aux[q[i]].
Below is the implementation of the above approach. 

C++

// C++ implementation of the above approach.
#include <bits/stdc++.h>

#define MAX 100002
using namespace std;

// Function to print no of unique elements
// in specified suffix for each query.
void printUniqueElementsInSuffix(const int* arr,
                                 int n, const int* q, int m)
{

    // aux[i] stores no of unique elements in arr[i..n]
    int aux[MAX];

    // mark[i] = 1 if i has occurred in the suffix at least once.
    int mark[MAX];
    memset(mark, 0, sizeof(mark));

    // Initialization for the last element.
    aux[n - 1] = 1;
    mark[arr[n - 1]] = 1;

    for (int i = n - 2; i >= 0; i--) {
        if (mark[arr[i]] == 0) {
            aux[i] = aux[i + 1] + 1;
            mark[arr[i]] = 1;
        }
        else {
            aux[i] = aux[i + 1];
        }
    }

    for (int i = 0; i < m; i++) {
        cout << aux[q[i] - 1] << "\n";
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 1, 2, 3, 4, 10000, 999 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int q[] = { 1, 3, 6 };
    int m = sizeof(q) / sizeof(q[0]);
    printUniqueElementsInSuffix(arr, n, q, m);
    return 0;
}

Java

// Java implementation of the above approach.
class GFG
{

static int MAX = 100002;

// Function to print no of unique elements
// in specified suffix for each query.
static void printUniqueElementsInSuffix(int[] arr,
                                int n, int[] q, int m)
{

    // aux[i] stores no of unique elements in arr[i..n]
    int []aux = new int[MAX];

    // mark[i] = 1 if i has occurred
    // in the suffix at least once.
    int []mark = new int[MAX];

    // Initialization for the last element.
    aux[n - 1] = 1;
    mark[arr[n - 1]] = 1;

    for (int i = n - 2; i >= 0; i--)
    {
        if (mark[arr[i]] == 0)
        {
            aux[i] = aux[i + 1] + 1;
            mark[arr[i]] = 1;
        }
        else
        {
            aux[i] = aux[i + 1];
        }
    }

    for (int i = 0; i < m; i++)
    {
        System.out.println(aux[q[i] - 1] );
    }
}

// Driver code
public static void main(String[] args)
{

    int arr[] = { 1, 2, 3, 4, 1, 2, 3, 4, 10000, 999 };
    int n = arr.length;
    int q[] = { 1, 3, 6 };
    int m = q.length;
    printUniqueElementsInSuffix(arr, n, q, m);
}
}

// This code is contributed by Princi Singh

Python3

# Python implementation of the above approach.
MAX = 100002;

# Function to print no of unique elements
# in specified suffix for each query.
def printUniqueElementsInSuffix(arr, n, q, m):

    # aux[i] stores no of unique elements in arr[i..n]
    aux = [0] * MAX;

    # mark[i] = 1 if i has occurred
    # in the suffix at least once.
    mark = [0] * MAX;

    # Initialization for the last element.
    aux[n - 1] = 1;
    mark[arr[n - 1]] = 1;

    for i in range(n - 2, -1, -1):
        if (mark[arr[i]] == 0):
            aux[i] = aux[i + 1] + 1;
            mark[arr[i]] = 1;
        else:
            aux[i] = aux[i + 1];

    for i in range(m):
        print(aux[q[i] - 1]);

# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 1, 2, 3, 4, 10000, 999];
    n = len(arr);
    q = [1, 3, 6];
    m = len(q);
    printUniqueElementsInSuffix(arr, n, q, m);

# This code is contributed by 29AjayKumar

C#

// C# implementation of the above approach.
using System;

class GFG
{
    static int MAX = 100002;

    // Function to print no of unique elements
    // in specified suffix for each query.
    static void printUniqueElementsInSuffix(int[] arr,
                                    int n, int[] q, int m)
    {

        // aux[i] stores no of unique elements in arr[i..n]
        int []aux = new int[MAX];

        // mark[i] = 1 if i has occurred
        // in the suffix at least once.
        int []mark = new int[MAX];

        // Initialization for the last element.
        aux[n - 1] = 1;
        mark[arr[n - 1]] = 1;

        for (int i = n - 2; i >= 0; i--)
        {
            if (mark[arr[i]] == 0)
            {
                aux[i] = aux[i + 1] + 1;
                mark[arr[i]] = 1;
            }
            else
            {
                aux[i] = aux[i + 1];
            }
        }

        for (int i = 0; i < m; i++)
        {
            Console.WriteLine(aux[q[i] - 1] );
        }
    }

    // Driver code
    static public void Main ()
    {

        int []arr = { 1, 2, 3, 4, 1, 2, 3, 4, 10000, 999 };
        int n = arr.Length;
        int []q = { 1, 3, 6 };
        int m = q.Length;
        printUniqueElementsInSuffix(arr, n, q, m);
    }
}

// This code is contributed by Sachin.

Javascript

<script>

// javascript implementation of the above approach.

     var MAX = 100002;

    // Function to print no of unique elements
    // in specified suffix for each query.
    function printUniqueElementsInSuffix(arr, n,  q,  m)
    {

        // aux[i] stores no of unique elements in arr[i..n]
        var aux =[] ;
        var mark = [];
        aux.length = MAX ;
        mark.length = MAX ;
        aux.fill(0);
        mark.fill(0);

        // Initialization for the last element.
        aux[n - 1] = 1;
        mark[arr[n - 1]] = 1;

        for (var i = n - 2; i >= 0; i--)
        {
            if (mark[arr[i]] == 0)
            {
                aux[i] = aux[i + 1] + 1;
                mark[arr[i]] = 1;
            }
            else
            {
                aux[i] = aux[i + 1];
            }
        }

        for (var i = 0; i < m; i++)
        {
            document.write(aux[q[i] - 1] , "<br>" );
        }
    }

    // Driver code

        var arr = [ 1, 2, 3, 4, 1, 2, 3, 4, 10000, 999 ]
        var n = arr.length;
        var q = [ 1, 3, 6 ];
        var m = q.length;
        printUniqueElementsInSuffix(arr, n, q, m);

        // This code is contributed by bunnyram19.
</script>

Output: 

6
6
5

Time Complexity: O(n + m). where n and m are the sizes of the given input arrays.
Auxiliary Space: O(MAX)