要添加的元素，以便数组中存在某个范围的所有元素

给定大小为N的数组。令A和B分别为数组中的最小值和最大值。任务是找到应该添加给给定数组的数字，以使[A，B]范围内的所有元素至少在数组中出现一次。

例子：

Input : arr[] = {4, 5, 3, 8, 6}
Output : 1
Only 7 to be added in the list.

Input : arr[] = {2, 1, 3}
Output : 0

方法1(排序)
1-对数组进行排序。
2-比较arr [i] == arr [i + 1] -1。如果不是，则更新计数= arr [i + 1] -arr [i] -1。
3-返回计数。

C++

// C++ program for above implementation
#include 
using namespace std;
  
// Function to count numbers to be added
int countNum(int arr[], int n)
{
    int count = 0;
  
    // Sort the array
    sort(arr, arr + n);
  
    // Check if elements are consecutive
    //  or not. If not, update count
    for (int i = 0; i < n - 1; i++)
        if (arr[i] != arr[i+1] && 
            arr[i] != arr[i + 1] - 1)
            count += arr[i + 1] - arr[i] - 1;
  
    return count;
}
  
// Drivers code
int main()
{
    int arr[] = { 3, 5, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countNum(arr, n) << endl;
    return 0;
}

Java

// java program for above implementation
import java.io.*;
import java.util.*;
  
public class GFG {
      
    // Function to count numbers to be added
    static int countNum(int []arr, int n)
    {
        int count = 0;
      
        // Sort the array
        Arrays.sort(arr);
      
        // Check if elements are consecutive
        // or not. If not, update count
        for (int i = 0; i < n - 1; i++)
            if (arr[i] != arr[i+1] && 
                arr[i] != arr[i + 1] - 1)
                count += arr[i + 1] - arr[i] - 1;
      
        return count;
    }
      
    // Drivers code
    static public void main (String[] args)
    {
          
        int []arr = { 3, 5, 8, 6 };
        int n = arr.length;
          
        System.out.println(countNum(arr, n));
    }
}
  
// This code is contributed by vt_m.

Python3

# python program for above implementation
  
# Function to count numbers to be added
def countNum(arr, n): 
      
    count = 0
  
    # Sort the array
    arr.sort()
  
    # Check if elements are consecutive
    # or not. If not, update count
    for i in range(0, n-1):
        if (arr[i] != arr[i+1] and
            arr[i] != arr[i + 1] - 1):
            count += arr[i + 1] - arr[i] - 1;
  
    return count
  
# Drivers code
arr = [ 3, 5, 8, 6 ]
n = len(arr)
print(countNum(arr, n))
  
# This code is contributed by Sam007

C#

// C# program for above implementation
using System;
  
public class GFG {
      
    // Function to count numbers to be added
    static int countNum(int []arr, int n)
    {
        int count = 0;
      
        // Sort the array
        Array.Sort(arr);
      
        // Check if elements are consecutive
        // or not. If not, update count
        for (int i = 0; i < n - 1; i++)
            if (arr[i] != arr[i+1] && 
                arr[i] != arr[i + 1] - 1)
                count += arr[i + 1] - arr[i] - 1;
      
        return count;
    }
      
    // Drivers code
    static public void Main ()
    {
          
        int []arr = { 3, 5, 8, 6 };
        int n = arr.Length;
          
        Console.WriteLine(countNum(arr, n));
    }
}
  
// This code is contributed by vt_m.

输出：

2

时间复杂度： O(n log n)

方法2(使用散列)
1-维护数组元素的哈希。
2-存储最小和最大元素。
3-从哈希中的最小到最大元素遍历
并计算元素是否不在哈希中。
4-返回计数。

C++

// C++ program for above implementation
#include 
using namespace std;
  
// Function to count numbers to be added
int countNum(int arr[], int n)
{
    unordered_set s;
    int count = 0, maxm = INT_MIN, minm = INT_MAX;
  
    // Make a hash of elements
    // and store minimum and maximum element
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);
        if (arr[i] < minm)
            minm = arr[i];
        if (arr[i] > maxm)
            maxm = arr[i];
    }
  
    // Traverse all elements from minimum
    // to maximum and count if it is not
    // in the hash
    for (int i = minm; i <= maxm; i++)
        if (s.find(arr[i]) == s.end())
            count++;
    return count;
}
  
// Drivers code
int main()
{
    int arr[] = { 3, 5, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countNum(arr, n) << endl;
    return 0;
}

Java

// Java implementation of the approach
import java.util.HashSet;
  
class GFG 
{
  
// Function to count numbers to be added
static int countNum(int arr[], int n)
{
    HashSet s = new HashSet<>();
    int count = 0, 
        maxm = Integer.MIN_VALUE, 
        minm = Integer.MAX_VALUE;
  
    // Make a hash of elements
    // and store minimum and maximum element
    for (int i = 0; i < n; i++) 
    {
        s.add(arr[i]);
        if (arr[i] < minm)
            minm = arr[i];
        if (arr[i] > maxm)
            maxm = arr[i];
    }
  
    // Traverse all elements from minimum
    // to maximum and count if it is not
    // in the hash
    for (int i = minm; i <= maxm; i++)
        if (!s.contains(i))
            count++;
    return count;
}
  
// Drivers code
public static void main(String[] args) 
{
    int arr[] = { 3, 5, 8, 6 };
    int n = arr.length;
    System.out.println(countNum(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Function to count numbers to be added
def countNum(arr, n):
  
    s = dict()
    count, maxm, minm = 0, -10**9, 10**9
  
    # Make a hash of elements and store 
    # minimum and maximum element
    for i in range(n):
        s[arr[i]] = 1
        if (arr[i] < minm):
            minm = arr[i]
        if (arr[i] > maxm):
            maxm = arr[i]
      
    # Traverse all elements from minimum
    # to maximum and count if it is not
    # in the hash
    for i in range(minm, maxm + 1):
        if i not in s.keys():
            count += 1
    return count
  
# Driver code
arr = [3, 5, 8, 6 ]
n = len(arr)
print(countNum(arr, n))
      
# This code is contributed by mohit kumar

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to count numbers to be added
static int countNum(int []arr, int n)
{
    HashSet s = new HashSet();
    int count = 0, 
        maxm = int.MinValue, 
        minm = int.MaxValue;
  
    // Make a hash of elements
    // and store minimum and maximum element
    for (int i = 0; i < n; i++) 
    {
        s.Add(arr[i]);
        if (arr[i] < minm)
            minm = arr[i];
        if (arr[i] > maxm)
            maxm = arr[i];
    }
  
    // Traverse all elements from minimum
    // to maximum and count if it is not
    // in the hash
    for (int i = minm; i <= maxm; i++)
        if (!s.Contains(i))
            count++;
    return count;
}
  
// Drivers code
public static void Main(String[] args) 
{
    int []arr = { 3, 5, 8, 6 };
    int n = arr.Length;
    Console.WriteLine(countNum(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

输出：

2

时间复杂度-O(max – min + 1)