检查字符串流中的重复项

给定一个包含公司员工姓名的字符串数组arr[] 。假设在一个系统中一个接一个地输入姓名，任务是检查当前姓名是否是第一次输入。
例子：

Input: arr[] = {“geeks”, “for”, “geeks”} 
Output: 
No 
No 
Yes
Input: arr[] = {“abc”, “aaa”, “cba”} 
Output: 
No 
No 
No 
 

方法：创建一个 unordered_set 来存储员工的姓名并开始遍历数组，如果当前姓名已经存在于集合中，则打印 Yes else 打印 No 并将其插入到集合中。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to insert the names
// and check whether they appear
// for the first time
void insertNames(string arr[], int n)
{
 
    // To store the names
    // of the employees
    unordered_set set;
    for (int i = 0; i < n; i++) {
 
        // If current name is appearing
        // for the first time
        if (set.find(arr[i]) == set.end()) {
            cout << "No\n";
            set.insert(arr[i]);
        }
        else {
            cout << "Yes\n";
        }
    }
}
 
// Driver code
int main()
{
    string arr[] = { "geeks", "for", "geeks" };
    int n = sizeof(arr) / sizeof(string);
 
    insertNames(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to insert the names
// and check whether they appear
// for the first time
static void insertNames(String arr[], int n)
{
 
    // To store the names
    // of the employees
    HashSet set = new HashSet();
    for (int i = 0; i < n; i++)
    {
 
        // If current name is appearing
        // for the first time
        if (!set.contains(arr[i]))
        {
            System.out.print("No\n");
            set.add(arr[i]);
        }
        else
        {
            System.out.print("Yes\n");
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    String arr[] = { "geeks", "for", "geeks" };
    int n = arr.length;
 
    insertNames(arr, n);
}
}
 
// This code contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
 
# Function to insert the names
# and check whether they appear
# for the first time
def insertNames(arr, n) :
 
    # To store the names
    # of the employees
    string = set();
     
    for i in range(n) :
 
        # If current name is appearing
        # for the first time
        if arr[i] not in string :
            print("No");
            string.add(arr[i]);
     
        else :
            print("Yes");
     
# Driver code
if __name__ == "__main__" :
 
    arr = [ "geeks", "for", "geeks" ];
    n = len(arr);
 
    insertNames(arr, n);
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to insert the names
// and check whether they appear
// for the first time
static void insertNames(String []arr, int n)
{
 
    // To store the names
    // of the employees
    HashSet set = new HashSet();
    for (int i = 0; i < n; i++)
    {
 
        // If current name is appearing
        // for the first time
        if (!set.Contains(arr[i]))
        {
            Console.Write("No\n");
            set.Add(arr[i]);
        }
        else
        {
            Console.Write("Yes\n");
        }
    }
}
 
// Driver code
public static void Main(String[] args)
{
    String []arr = { "geeks", "for", "geeks" };
    int n = arr.Length;
 
    insertNames(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
    // Javascript implementation of the approach
     
// Function to insert the names
// and check whether they appear
// for the first time
function insertNames(arr, n)
{
   
    // To store the names
    // of the employees
    let set = new Set();
    for (let i = 0; i < n; i++)
    {
   
        // If current name is appearing
        // for the first time
        if (!set.has(arr[i]))
        {
            document.write("No"  + "<br/>");
            set.add(arr[i]);
        }
        else
        {
            document.write("Yes" + "<br/>");
        }
    }
}
     
    // Driver code
     
    let arr = [ "geeks", "for", "geeks" ];
    let n = arr.length;
   
    insertNames(arr, n);
  
 // This code is contributed by susmitakundugoaldanga.

输出：

No
No
Yes

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