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 Kth element to Nth 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 allows 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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)
  
    # Pr 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right


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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Output:

6
6
5


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.