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

给定一个大小为 n 的数组 ‘a[]’ 和查询数 q。每个查询都可以用一个整数 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：

• 以数组check[]为例，它会检查当前元素是否较早被访问过。如果已经访问过将其标记为1否则为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。
• 从最后一个循环，如果当前元素未被访问，则将其检查标记为1 ，将当前计数器存储在idx 中并将其递增，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 increment 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 increment 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 increment 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 increment 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 */

Javascript

 
 
// JavaScript implementation of the approach
 
// Function to perform queries to find
// number of distinct elements from
// a given index till last index in an array
function find_distinct(a, n, q, queries)
{
    let MAX =100001;
    let check = new Array(MAX).fill(0);
    let idx = new Array(MAX).fill(0);
    let cnt = 1;
     
    let i=n-1;
     
    while (i>=0) {
     // Check if current element
        // already visited or not
        if (check[a[i]] == 0)
        {
 
            // If not visited store current counter
            // and increment 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;
        }
    i--;
    }
     
 
    // Perform queries
    for (let i = 0; i < q; i++)
    {
            let m = queries[i];
            document.write(idx[m] + " ");
    }
     
}
 
// Driver code
 
let a = [1, 2, 3, 1, 2, 3, 4, 5 ];
let n = a.length;
let queries = [0, 3, 5, 7];
let q = queries.length;
find_distinct(a, n, q, queries);
 
 
 

输出：

5 5 3 1

时间复杂度： O(MAX + N)
辅助空间： O(MAX)