查询从给定索引到数组中最后一个索引的不同元素的数量

给定大小为n且查询数量为q的数组’a []’。每个查询可以用整数m表示。您的任务是打印从索引m到n的不同整数的数量，即直到数组的最后一个元素。

例子：

Input: arr[] = {1, 2, 3, 1, 2, 3, 4, 5}, q[] = {1, 4, 6, 8}
Output: 5 5 3 1
In query 1, number of distinct integers
in a[0…7] is 5 (1, 2, 3, 4, 5)
In query 2, number of distinct integers
in a[3…7] is 5 (1, 2, 3, 4, 5)
In query 3, number of distinct integers
in a[5…7] is 3 (3, 4, 5)
In query 4, number of distinct integers
in a[7…7] is 1 (5)

方法：

• 采取数组check [] ，它将检查是否较早访问了当前元素。如果已经访问过，则将其标记为1，否则将其标记为0 。
• 取一个数组idx [] ，它将存储从当前索引到最后一个索引的不同元素的数量。
• 从最后开始循环，如果尚未访问当前元素，则将其检查标记为1 ，将当前计数器存储在idx中，然后递增它，否则只需将当前计数器存储在idx中。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define MAX 100001
  
// Function to perform queries to find
// number of distinct elements from
// a given index till last index in an array
void find_distinct(int a[], int n, int q, int queries[])
{
    int check[MAX] = { 0 };
    int idx[MAX];
    int cnt = 1;
    for (int i = n - 1; i >= 0; i--) {
        // Check if current element
        // already visited or not
        if (check[a[i]] == 0) {
  
            // If not visited store current counter
            // and incremet it and mark check as 1
            idx[i] = cnt;
            check[a[i]] = 1;
            cnt++;
        }
        else {
  
            // Otherwise if visited simply
            // store current counter
            idx[i] = cnt - 1;
        }
    }
  
    // Perform queries
    for (int i = 0; i < q; i++) {
        int m = queries[i];
        cout << idx[m] << " ";
    }
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 3, 1, 2, 3, 4, 5 };
    int n = sizeof(a) / sizeof(int);
    int queries[] = { 0, 3, 5, 7 };
    int q = sizeof(queries) / sizeof(int);
    find_distinct(a, n, q, queries);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
static int MAX =100001; 
  
// Function to perform queries to find 
// number of distinct elements from 
// a given index till last index in an array 
static void find_distinct(int a[], int n, int q, int queries[]) 
{ 
    int []check = new int[MAX]; 
    int []idx = new int[MAX]; 
    int cnt = 1; 
    for (int i = n - 1; i >= 0; i--) 
    { 
        // Check if current element 
        // already visited or not 
        if (check[a[i]] == 0) 
        { 
  
            // If not visited store current counter 
            // and incremet it and mark check as 1 
            idx[i] = cnt; 
            check[a[i]] = 1; 
            cnt++; 
        } 
        else 
        { 
  
            // Otherwise if visited simply 
            // store current counter 
            idx[i] = cnt - 1; 
        } 
    } 
  
    // Perform queries 
    for (int i = 0; i < q; i++)
    { 
            int m = queries[i]; 
            System.out.print(idx[m] + " "); 
    } 
} 
  
// Driver code 
public static void main(String[] args) 
{
    int a[] = { 1, 2, 3, 1, 2, 3, 4, 5 }; 
    int n = a.length; 
    int queries[] = { 0, 3, 5, 7 }; 
    int q = queries.length; 
    find_distinct(a, n, q, queries);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python implementation of the approach
MAX = 100001;
  
# Function to perform queries to find
# number of distinct elements from
# a given index till last index in an array
def find_distinct(a, n, q, queries):
    check = [0] * MAX;
    idx = [0] * MAX;
    cnt = 1;
    for i in range(n - 1, -1, -1):
          
        # Check if current element
        # already visited or not
        if (check[a[i]] == 0):
  
            # If not visited store current counter
            # and incremet it and mark check as 1
            idx[i] = cnt;
            check[a[i]] = 1;
            cnt += 1;
        else:
  
            # Otherwise if visited simply
            # store current counter
            idx[i] = cnt - 1;
  
    # Perform queries
    for i in range(0, q):
        m = queries[i];
        print(idx[m], end = " ");
  
# Driver code
a = [ 1, 2, 3, 1, 2, 3, 4, 5 ];
n = len(a);
queries = [ 0, 3, 5, 7 ];
q = len(queries);
find_distinct(a, n, q, queries);
  
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System; 
      
class GFG 
{
      
static int MAX =100001; 
  
// Function to perform queries to find 
// number of distinct elements from 
// a given index till last index in an array 
static void find_distinct(int []a, int n, int q, int []queries) 
{ 
    int []check = new int[MAX]; 
    int []idx = new int[MAX]; 
    int cnt = 1; 
    for (int i = n - 1; i >= 0; i--) 
    { 
        // Check if current element 
        // already visited or not 
        if (check[a[i]] == 0) 
        { 
  
            // If not visited store current counter 
            // and incremet it and mark check as 1 
            idx[i] = cnt; 
            check[a[i]] = 1; 
            cnt++; 
        } 
        else
        { 
  
            // Otherwise if visited simply 
            // store current counter 
            idx[i] = cnt - 1; 
        } 
    } 
  
    // Perform queries 
    for (int i = 0; i < q; i++)
    { 
            int m = queries[i]; 
            Console.Write(idx[m] + " "); 
    } 
} 
  
// Driver code 
public static void Main(String[] args) 
{
    int []a = { 1, 2, 3, 1, 2, 3, 4, 5 }; 
    int n = a.Length; 
    int []queries = { 0, 3, 5, 7 }; 
    int q = queries.Length; 
    find_distinct(a, n, q, queries);
}
}
  
/* This code is contributed by PrinciRaj1992 */

输出：

5 5 3 1