Skip to content
Related Articles

Queries for number of distinct integers in Suffix

• Difficulty Level : Easy
• Last Updated : 02 Jun, 2021

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 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 ``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 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

 `= 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

 ``
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[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 prno 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

 ``
Output:
```6
6
5```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up