未排序数组中的第 k 个缺失元素

给定一个未排序的序列 a[]，任务是在数组元素的递增序列中找到第 K 个缺失的连续元素，即按排序顺序考虑数组并找到第 k 个缺失的数字。如果没有第 k 个缺失元素，则输出 -1。
注意：只有元素存在于要考虑的最小和最大元素范围内。
例子：

Input: arr[] = {2, 4, 10, 7}, k = 5
Output: 9
Missing elements in the given array: 3, 5, 6, 8, 9
5th missing is 9.

Input: arr[] = {1, 3, 4}, k = 5
Output: -1

方法 1：对数组进行排序，并使用已排序数组中第 k 个缺失元素中使用的方法。
方法 2：

1. 将所有元素插入到 unordered_set 中。
2. 找到数组的最小和最大元素。
3. 从最小到最大遍历元素。
   • 检查当前元素是否存在于集合中。
   • 如果不是，则通过计算丢失的元素来检查这是否是第 k 个丢失。
   • 如果当前缺少当前元素，则返回当前元素。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to find the sum
// of minimum of all subarrays
int findKth(int arr[], int n, int k)
{
 
    unordered_set missing;
    int count = 0;
 
    // Insert all the elements in a set
    for (int i = 0; i < n; i++)
        missing.insert(arr[i]);
 
    // Find the maximum and minimum element
    int maxm = *max_element(arr, arr + n);
    int minm = *min_element(arr, arr + n);
 
    // Traverse from the minimum to maximum element
    for (int i = minm + 1; i < maxm; i++) {
        // Check if "i" is missing
        if (missing.find(i) == missing.end())
            count++;
 
        // Check if it is kth missing
        if (count == k)
            return i;
    }
 
    // If no kth element is missing
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 10, 9, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 5;
    cout << findKth(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
 
    // Function to find the sum
    // of minimum of all subarrays
    static int findKth(int arr[], int n, int k)
    {
 
        HashSet missing = new HashSet<>();
        int count = 0;
 
        // Insert all the elements in a set
        for (int i = 0; i < n; i++)
        {
            missing.add(arr[i]);
        }
 
        // Find the maximum and minimum element
        int maxm = Arrays.stream(arr).max().getAsInt();
        int minm = Arrays.stream(arr).min().getAsInt();
 
        // Traverse from the minimum to maximum element
        for (int i = minm+1; i < maxm; i++)
        {
            // Check if "i" is missing
            if (!missing.contains(i))
            {
                count++;
            }
 
            // Check if it is kth missing
            if (count == k)
            {
                return i;
            }
        }
         
        // If no kth element is missing
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {2, 10, 9, 4};
        int n = arr.length;
        int k = 5;
        System.out.println(findKth(arr, n, k));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python

# Python3 implementation of the above approach
 
# Function to find the sum
# of minimum of all subarrays
def findKth( arr, n, k):
 
    missing = dict()
    count = 0
 
    # Insert all the elements in a set
    for i in range(n):
        missing[arr[i]] = 1
 
    # Find the maximum and minimum element
    maxm = max(arr)
    minm = min(arr)
 
    # Traverse from the minimum to maximum element
    for i in range(minm + 1, maxm):
         
        # Check if "i" is missing
        if (i not in missing.keys()):
            count += 1
 
        # Check if it is kth missing
        if (count == k):
            return i
     
    # If no kth element is missing
    return -1
 
# Driver code
arr = [2, 10, 9, 4 ]
n = len(arr)
k = 5
print(findKth(arr, n, k))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the above approach
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find the sum
    // of minimum of all subarrays
    static int findKth(int []arr, int n, int k)
    {
 
        HashSet missing = new HashSet();
        int count = 0;
 
        // Insert all the elements in a set
        for (int i = 0; i < n; i++)
        {
            missing.Add(arr[i]);
        }
 
        // Find the maximum and minimum element
        int maxm = arr.Max();
        int minm = arr.Min();
 
        // Traverse from the minimum to maximum element
        for (int i = minm + 1; i < maxm; i++)
        {
            // Check if "i" is missing
            if (!missing.Contains(i))
            {
                count++;
            }
 
            // Check if it is kth missing
            if (count == k)
            {
                return i;
            }
        }
         
        // If no kth element is missing
        return -1;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {2, 10, 9, 4};
        int n = arr.Length;
        int k = 5;
        Console.WriteLine(findKth(arr, n, k));
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript

 
// javascript implementation of the above approach
 
// Function to find the sum
// of minimum of all subarrays
function findKth(arr, n, k)
{
 
    var missing = new Set();
    var count = 0;
 
    // Insert all the elements in a set
    for (var i = 0; i < n; i++)
        missing.add(arr[i]);
 
    // Find the maximum and minimum element
    var maxm = arr.reduce((a,b)=>Math.max(a,b));
    var minm = arr.reduce((a,b)=>Math.min(a,b));
 
    // Traverse from the minimum to maximum element
    for (var i = minm + 1; i < maxm; i++) {
        // Check if "i" is missing
        if (!missing.has(i))
            count++;
 
        // Check if it is kth missing
        if (count == k)
            return i;
    }
 
    // If no kth element is missing
    return -1;
}
 
// Driver code
var arr = [ 2, 10, 9, 4 ];
var n = arr.length;
var k = 5;
document.write( findKth(arr, n, k));
 
// This code is contributed by noob2000.

输出：

8

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。