数组中的高尚整数(大于等于的元素数等于value)

给定数组arr []，在其中找到一个Noble整数。如果大于x的整数数量等于x，则在arr []中将整数x称为Noble。如果有许多Noble整数，则返回其中的任何一个。如果没有，则返回-1。
例子 ：

Input  : [7, 3, 16, 10]
Output : 3  
Number of integers greater than 3
is three.

Input  : [-1, -9, -2, -78, 0]
Output : 0
Number of integers greater than 0
is zero.

方法1(蛮力)
遍历数组。对于每个元素arr [i]，找到大于arr [i]的元素数量。

CPP

// C++ program to find Noble elements
// in an array.
#include 
using namespace std;
 
// Returns a Noble integer if present,
// else returns -1.
int nobleInteger(int arr[], int size)
{
    for (int i = 0; i < size; i++ )
    {
        int count = 0;
        for (int j = 0; j < size; j++)
            if (arr[i] < arr[j])
                count++;
                 
        // If count of greater elements
        // is equal to arr[i]
        if (count == arr[i])
            return arr[i];
    }
     
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = {10, 3, 20, 40, 2};
    int size = sizeof(arr) / sizeof(arr[0]);
    int res = nobleInteger(arr, size);
     
    if (res != -1)
        cout<<"The noble integer is "<< res;
    else
        cout<<"No Noble Integer Found";
}
 
// This code is contributed by Smitha.

Java

// Java program to find Noble elements
// in an array.
import java.util.ArrayList;
 
class GFG {
     
    // Returns a Noble integer if present,
    // else returns -1.
    public static int nobleInteger(int arr[])
    {
        int size = arr.length;
        for (int i = 0; i < size; i++ )
        {
            int count = 0;
            for (int j = 0; j < size; j++)
                if (arr[i] < arr[j])
                    count++;
 
            // If count of greater elements
            // is equal to arr[i]
            if (count == arr[i])
                return arr[i];
        }
        return -1;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int [] arr = {10, 3, 20, 40, 2};
        int res = nobleInteger(arr);
        if (res != -1)
            System.out.println("The noble "
                     + "integer is "+ res);
        else
            System.out.println("No Noble "
                        + "Integer Found");
    }
}

Python3

# Python3 program to find Noble
# elements in an array.
 
# Returns a Noble integer if
# present, else returns -1.
def nobleInteger(arr, size):
 
    for i in range(0, size):
     
        count = 0
        for j in range(0, size):
            if (arr[i] < arr[j]):
                count += 1
        # If count of greater
        # elements is equal
        # to arr[i]
        if (count == arr[i]):
            return arr[i]
     
    return -1
 
# Driver code
arr = [10, 3, 20, 40, 2]
size = len(arr)
res = nobleInteger(arr,size)
if (res != -1):
    print("The noble integer is ",
                              res)
else:
    print("No Noble Integer Found")
 
# This code is contributed by
# Smitha.

C#

// C# program to find Noble elements
// in an array.
using System;
 
class GFG {
     
    // Returns a Noble integer if present,
    // else returns -1.
    public static int nobleInteger(int [] arr)
    {
        int size = arr.Length;
        for (int i = 0; i < size; i++ )
        {
            int count = 0;
            for (int j = 0; j < size; j++)
                if (arr[i] < arr[j])
                    count++;
 
            // If count of greater elements
            // is equal to arr[i]
            if (count == arr[i])
                return arr[i];
        }
         
        return -1;
    }
 
    // Driver code
    public static void Main()
    {
        int [] arr = {10, 3, 20, 40, 2};
        int res = nobleInteger(arr);
        if (res != -1)
            Console.Write("The noble integer"
                              + " is "+ res);
        else
            Console.Write("No Noble Integer"
                                 + " Found");
    }
}
 
// This code is contributed by Smitha.

---

输出

The noble integer is 3

方法2(使用排序)

1. 以升序对数组arr []进行排序。此步骤需要(O(nlogn))。
2. 遍历数组。将索引i的值与索引i之后的元素数进行比较。如果arr [i]等于arr [i]之后的元素数，则它是一个高尚的整数。检查条件：(A [i] == length-i-1)。此步骤需要O(n)。

注意：数组可能具有重复的元素。因此，我们应该跳过相同的元素(排序数组中的相邻元素)。

C++

// C++ program to find Noble elements
// in an array.
#include
using namespace std;
 
// Returns a Noble integer if present,
// else returns -1.
int nobleInteger(int arr[], int n)
{
    sort(arr, arr + n);
 
    // Return a Noble element if present
    // before last.
    for (int i = 0; i < n - 1; i++)
    {
        if (arr[i] == arr[i + 1])
            continue;
 
        // In case of duplicates, we
        // reach last occurrence here.
        if (arr[i] == n - i - 1)
            return arr[i];
    }
    if (arr[n - 1] == 0)
        return arr[n - 1];
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = {10, 3, 20, 40, 2};
    int res = nobleInteger(arr, 5);
    if (res != -1)
        cout << "The noble integer is " << res;
    else
        cout << "No Noble Integer Found";
    return 0;
}
 
// This code is contributed by Rajput-Ji

Java

// Java program to find Noble elements
// in an array.
import java.util.Arrays;
 
public class Main
{
    // Returns a Noble integer if present,
    // else returns -1.
    public static int nobleInteger(int arr[])
    {
        Arrays.sort(arr);
 
        // Return a Noble element if present
        // before last.
        int n = arr.length;
        for (int i=0; i

Python

# Python3 code to find Noble elements
# in an array
 
def nobleInteger(arr):
     
    arr.sort()
     
    # Return a Noble element if
    # present before last
    n = len(arr)
     
    for i in range(n - 1):
         
        if arr[i] == arr[i + 1]:
            continue
             
        # In case of duplicates we reach
        # last occurrence here
        if arr[i] == n - i - 1:
            return arr[i]
     
    if arr[n - 1] == 0:
        return arr[n - 1]
    return -1
 
# Driver code
arr = [10, 3, 20, 40, 2]
 
res = nobleInteger(arr)
 
if res != -1:
    print("The noble integer is", res)
else:
    print("No Noble Integer Found")
 
# This code is contributed
# by Mohit Kumar

C＃

// C# program to find Noble elements
// in an array.
using System;
 
public class GFG {
     
    public static int nobleInteger(int[] arr)
    {
        Array.Sort(arr);
 
        // Return a Noble element if present
        // before last.
        int n = arr.Length;
        for (int i = 0; i < n-1; i++)
        {
            if (arr[i] == arr[i+1])
                continue;
 
            // In case of duplicates, we
            // reach last occurrence here.
            if (arr[i] == n-i-1)
                return arr[i];
        }
 
        if (arr[n-1] == 0)
            return arr[n-1];
 
        return -1;
    }
     
    // Driver code
    static public void Main ()
    {
        int [] arr = {10, 3, 20, 40, 2};
        int res = nobleInteger(arr);
        if (res != -1)
        Console.Write("The noble integer is "
                                      + res);
        else
            Console.Write("No Noble Integer "
                                  + "Found");
         
    }
}
 
// This code is contributed by Shrikant13.

输出

The noble integer is 3

方法3(使用计数数组)

保持一个计数数组countArr []，使所有元素的计数都大于或等于arr [i]。

1. 声明一个大小为n + 1(其中n是给定数组arr的大小)的整数数组countArr []，并将其初始化为零。
2. 遍历数组arr，如果arr [i] <0，则忽略它，如果arr [i]> = n，则递增countArr [n]，否则仅递增countArr [arr [i]]。
3. 声明一个整数totalGreater，该元素使元素数大于当前元素数，并将其初始化为countArr [arr [n]]。
4. 从最后一个索引到第一个索引遍历count数组countArr，如果在任何时候我们发现对于countArr [i]> 0，totalGreater = i，我们就找到了解决方案。否则，继续使用countArr [i]来增加totalGreater。

C++

// C++ program to find Noble elements
// in an array.
#include 
using namespace std;
 
int nobleInteger(int arr[], int n)
{
     
    // Declare a countArr which keeps
    // count of all elements
    // greater than or equal to arr[i].
    // Initialize it with zero.
    int countArr[n + 1] = { 0 };
 
    // Iterating through the given array
    for (int i = 0; i < n; i++) {
       
        // If current element is less
        // than zero, it cannot
        // be a solution so we skip it.
        if (arr[i] < 0) {
            continue;
        }
 
        // If current element is >= size of input array, if
        // will be greater than all elements which can be
        // considered as our solution, as it cannot be
        // greater than size of array.
        else if (arr[i] >= n) {
            countArr[n]++;
        }
 
        // Else we increase the count
        // of elements >= our
        // current array in countArr
        else {
            countArr[arr[i]]++;
        }
    }
 
    // Initially, countArr[n] is
    // count of elements greater
    // than all possible solutions
    int totalGreater = countArr[n];
 
    // Iterating through countArr
    for (int i = n - 1; i >= 0; i--) {
       
        // If totalGreater = current index, means we found
        // arr[i] for which count of elements >= arr[i] is
        // equal to arr[i]
        if (totalGreater == i && countArr[i] > 0) {
            return i;
        }
       
        // If at any point count of elements greater than
        // arr[i] becomes more than current index, then it
        // means we can no longer have a solution
        else if (totalGreater > i) {
            return -1;
        }
 
        // Adding count of elements >= arr[i] to
        // totalGreater.
        totalGreater += countArr[i];
    }
 
    return -1;
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 3, 20, 40, 2 };
    int res = nobleInteger(arr, 5);
    if (res != -1)
        cout << "The noble integer is " << res;
    else
        cout << "No Noble Integer Found";
    return 0;
}

输出

The noble integer is 3

• 复杂度分析：
  • 时间复杂度：O(n)。当我们遍历输入数组和countArr一次时。
  • 空间复杂度：O(n)。由于我们使用的countArr大小与给定数组相同。