数组中的第 k 个不同（或非重复）元素。

给定一个整数数组，打印数组中的第 k 个不同元素。给定的数组可能包含重复项，输出应打印所有唯一元素中的第 k 个元素。如果 k 大于不同元素的数量，则打印 -1。
例子 ：

Input : arr[] = {1, 2, 1, 3, 4, 2}, 
        k = 2
Output : 4

First non-repeating element is 3
Second non-repeating element is 4

Input : arr[] = {1, 2, 50, 10, 20, 2}, 
        k = 3
Output : 10

Input : {2, 2, 2, 2}, 
        k = 2
Output : -1

一个简单的解决方案是使用两个嵌套循环，其中外循环从左到右选取元素，内循环检查选取的元素是否存在于其他地方。如果不存在，则增加不同元素的计数。如果 count 变为 k，则返回当前元素。

C++

// C++ program to print k-th distinct
// element in a given array
#include 
using namespace std;
 
// Returns k-th distinct
// element in arr.
int printKDistinct(int arr[], int n,
                              int k)
{
    int dist_count = 0;
    for (int i = 0; i < n; i++)
    {
        // Check if current element is
        // present somewhere else.
        int j;
        for (j = 0; j < n; j++)
            if (i != j && arr[j] == arr[i])
                break;
 
        // If element is unique
        if (j == n)
            dist_count++;
 
        if (dist_count == k)
            return arr[i];
    }
 
    return -1;
}
 
// Driver Code
int main ()
{
    int ar[] = {1, 2, 1, 3, 4, 2};
    int n = sizeof(ar) / sizeof(ar[0]);
    int k = 2;
    cout << printKDistinct(ar, n, k);
    return 0;
}

Java

// Java program to print k-th distinct
// element in a given array
class GFG
{
    // Returns k-th distinct element in arr.
    static int printKDistinct(int arr[],
                                  int n,
                                  int k)
    {
        int dist_count = 0;
        for (int i = 0; i < n; i++)
        {
             
            // Check if current element is
            // present somewhere else.
            int j;
             
            for (j = 0; j < n; j++)
                if (i != j && arr[j] == arr[i])
                    break;
     
            // If element is unique
            if (j == n)
                dist_count++;
     
            if (dist_count == k)
                return arr[i];
        }
     
        return -1;
    }
     
    //Driver code
    public static void main (String[] args)
    {
         
        int ar[] = {1, 2, 1, 3, 4, 2};
        int n = ar.length;
        int k = 2;
         
        System.out.print(printKDistinct(ar, n, k));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to prk-th distinct
# element in a given array
 
# Returns k-th distinct
# element in arr.
def printKDistinct(arr, n, k):
    dist_count = 0
    for i in range(n):
         
        # Check if current element is
        # present somewhere else.
        j = 0
        while j < n:
            if (i != j and arr[j] == arr[i]):
                break
            j += 1
 
        # If element is unique
        if (j == n):
            dist_count += 1
 
        if (dist_count == k):
            return arr[i]
 
    return -1
 
# Driver Code
ar = [1, 2, 1, 3, 4, 2]
n = len(ar)
k = 2
print(printKDistinct(ar, n, k))
 
# This code is contributed by Mohit Kumar

C#

// C# program to print k-th distinct
// element in a given array
using System;
 
class GFG
{
    // Returns k-th distinct element in arr
    static int printKDistinct(int []arr,
                                  int n,
                                  int k)
    {
         
        int dist_count = 0;
        for (int i = 0; i < n; i++)
        {
             
            // Check if current element is
            // present somewhere else.
            int j;
             
            for (j = 0; j < n; j++)
                if (i != j && arr[j] == arr[i])
                    break;
     
            // If element is unique
            if (j == n)
                dist_count++;
     
            if (dist_count == k)
                return arr[i];
        }
     
        return -1;
    }
     
    //Driver code
    public static void Main ()
    {
         
        int []ar = {1, 2, 1, 3, 4, 2};
        int n = ar.Length;
        int k = 2;
         
        Console.Write(printKDistinct(ar, n, k));
    }
}
 
// This code is contributed by nitn mittal

Javascript

//Javascript program to print k-th distinct
// element in a given array
 
// Returns k-th distinct
// element in arr.
function printKDistinct(arr, n,  k)
{
    var dist_count = 0;
    for (var i = 0; i < n; i++)
    {
        // Check if current element is
        // present somewhere else.
        var j;
        for (j = 0; j < n; j++)
            if (i != j && arr[j] == arr[i])
                break;
 
        // If element is unique
        if (j == n)
            dist_count++;
 
        if (dist_count == k)
            return arr[i];
    }
 
    return -1;
}
 
var ar = [1, 2, 1, 3, 4, 2];
    var n = ar.length;
    var k = 2;
    document.write( printKDistinct(ar, n, k));
 
//This code is contributed by SoumikMondal

输出 ：

4

一个有效的解决方案是使用哈希在平均 O(n) 时间内解决这个问题。
1)创建一个空的哈希表。
2）从左到右遍历输入数组并将元素及其计数存储在哈希表中。
3)再次从左到右遍历输入数组。继续计数计数为 1 的元素。
4)如果count 变为k，则返回当前元素。

C++

// C++ program to print k-th
// distinct element in a
// given array
#include 
using namespace std;
 
// Returns k-th distinct
// element in arr
int printKDistinct(int arr[],
                   int n, int k)
{
    // Traverse input array and
    // store counts if individual
    // elements.
    unordered_map h;
    for (int i = 0; i < n; i++)
        h[arr[i]]++;
 
    // If size of hash is
    // less than k.
    if (h.size() < k)
        return -1;
 
    // Traverse array again and
    // find k-th element with
    // count as 1.
    int dist_count = 0;
    for (int i = 0; i < n; i++)
    {
        if (h[arr[i]] == 1)
            dist_count++;
        if (dist_count == k)
            return arr[i];
    }
 
    return -1;
}
 
// Driver Code
int main ()
{
    int ar[] = {1, 2, 1, 3, 4, 2};
    int n = sizeof(ar) / sizeof(ar[0]);
    cout << printKDistinct(ar, n, 2);
    return 0;
}

Java

// Java program to print k-th distinct
// element in a given array
import java.util.*;
 
class GfG
{
 
// Returns k-th distinct
// element in arr.
static int printKDistinct(int arr[],
                        int n, int k)
{
    //int dist_count = 0;
    Map  h =
       new HashMap ();
        
    for (int i = 0; i < n; i++)
    {
        if(h.containsKey(arr[i]))
            h.put(arr[i], h.get(arr[i]) + 1);
        else
            h.put(arr[i], 1);
    }
 
    // If size of hash is
    // less than k.
    if (h.size() < k)
        return -1;
 
    // Traverse array again and
    // find k-th element with
    // count as 1.
    int dist_count = 0;
    for (int i = 0; i < n; i++)
    {
        if (h.get(arr[i]) == 1)
            dist_count++;
        if (dist_count == k)
            return arr[i];
    }
    return -1;
}
 
// Driver Code
public static void main (String[] args)
{
    int ar[] = {1, 2, 1, 3, 4, 2};
    int n = ar.length;
    System.out.println(printKDistinct(ar, n, 2));
}
}
 
// This code is contributed by
// Prerna Saini

蟒蛇3

# Python3 program to print k-th
# distinct element in a given array
def printKDistinct(arr, size, KthIndex):
    dict = {}
    vect = []
    for i in range(size):
        if(arr[i] in dict):
            dict[arr[i]] = dict[arr[i]] + 1
        else:
            dict[arr[i]] = 1
    for i in range(size):
        if(dict[arr[i]] > 1):
            continue
        else:
            KthIndex = KthIndex - 1
        if(KthIndex == 0):
            return arr[i]
    return -1
 
# Driver Code
arr = [1, 2, 1, 3, 4, 2]
size = len(arr)
print(printKDistinct(arr, size, 2))
 
# This code is contributed
# by Akhand Pratap Singh

C＃

// C# program to print k-th distinct
// element in a given array
using System;
using System.Collections.Generic;
 
class GfG
{
 
// Returns k-th distinct
// element in arr.
static int printKDistinct(int []arr,
                        int n, int k)
{
    Dictionary h = new Dictionary();
    for (int i = 0; i < n; i++)
    {
        if(h.ContainsKey(arr[i]))
        {
            var val = h[arr[i]];
            h.Remove(arr[i]);
            h.Add(arr[i], val + 1);
             
        }    
        else
            h.Add(arr[i], 1);
    }
     
    // If size of hash is
    // less than k.
    if (h.Count < k)
        return -1;
 
    // Traverse array again and
    // find k-th element with
    // count as 1.
    int dist_count = 0;
    for (int i = 0; i < n; i++)
    {
        if (h[arr[i]] == 1)
            dist_count++;
        if (dist_count == k)
            return arr[i];
    }
    return -1;
}
 
// Driver Code
public static void Main (String[] args)
{
    int []ar = {1, 2, 1, 3, 4, 2};
    int n = ar.Length;
    Console.WriteLine(printKDistinct(ar, n, 2));
}
}
 
// This code contributed by Rajput-Ji

Javascript

// Javascript program to print k-th distinct
// element in a given array
     
    // Returns k-th distinct
    // element in arr.
    function printKDistinct(arr,n,k)
    {
        // int dist_count = 0;
    let h = new Map();
         
    for (let i = 0; i < n; i++)
    {
        if(h.has(arr[i]))
            h.set(arr[i], h.get(arr[i]) + 1);
        else
            h.set(arr[i], 1);
    }
  
    // If size of hash is
    // less than k.
    if (h.length < k)
        return -1;
  
    // Traverse array again and
    // find k-th element with
    // count as 1.
    let dist_count = 0;
    for (let i = 0; i < n; i++)
    {
        if (h.get(arr[i]) == 1)
            dist_count++;
        if (dist_count == k)
            return arr[i];
    }
    return -1;
    }
     
    // Driver Code
    let ar=[1, 2, 1, 3, 4, 2];
    let n = ar.length;
    document.write(printKDistinct(ar, n, 2));
     
// This code is contributed by unknown2108

输出 ：

4

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