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

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

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++

 // C++ program to find distinct // elements in suffix    #include 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 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 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)        # 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

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 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.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

 = 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 ?>

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    #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);     int q[] = { 1, 3, 6 };     int m = sizeof(q) / sizeof(q);     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

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.

Output:

6
6
5

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