检查数组是否仅包含一个不同的元素

给定大小为N的数组arr [] ，任务是检查该数组是否仅包含一个不同的元素。如果它仅包含一个不同的元素，则打印“是” ，否则打印“否” 。

例子：

Input: arr[] = {3, 3, 4, 3, 3} 
Output: No  
Explanation: 
There are 2 distinct elements present in the array {3, 4}. 
Therefore, the output is No.

Input: arr[] = {9, 9, 9, 9, 9, 9, 9} 
Output: Yes 
Explanation: 
The only distinct element in the array is 9. 
Therefore, the output is Yes.

天真的方法：想法是对给定的数组进行排序，然后对每个有效索引检查当前元素和下一个元素是否相同。如果它们不相同，则表示该数组包含多个不同的元素，因此打印“否” ，否则打印“是” 。

时间复杂度： O(N * logN)
辅助空间： O(1)

更好的方法：可以通过使用一组数据结构来解决此问题。自设置以来，不允许重复。步骤如下：

1. 将数组的元素插入集合。
2. 如果只有一个不同的元素，则步骤1之后的集合的大小将为1，因此打印“是” 。
3. 否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find if the array
// contains only one distinct element
void uniqueElement(int arr[],int n)
{
     
    // Create a set
    unordered_set set;
     
    // Traversing the array
    for(int i = 0; i < n; i++)
    {
        set.insert(arr[i]);
    }
     
    // Compare and print the result
    if(set.size() == 1)
    {
        cout << "YES" << endl;
    }
    else
    {
        cout << "NO" << endl;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 9, 9, 9, 9, 9, 9, 9 };
     
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    uniqueElement(arr,n);
    return 0;
}
 
// This code is contributed by rutvik_56

Java

// Java program for the above approach
import java.util.*;
 
public class Main {
 
    // Function to find if the array
    // contains only one distinct element
    public static void
    uniqueElement(int arr[])
    {
        // Create a set
        Set set = new HashSet<>();
 
        // Traversing the array
        for (int i = 0; i < arr.length; i++) {
            set.add(arr[i]);
        }
 
        // Compare and print the result
        if (set.size() == 1)
            System.out.println("Yes");
 
        else
            System.out.println("No");
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 9, 9, 9, 9, 9, 9, 9 };
 
        // Function call
        uniqueElement(arr);
    }
}

Python3

# Python3 program for the above approach
 
# Function to find if the array
# contains only one distinct element
def uniqueElement(arr, n):
     
    # Create a set
    s = set(arr)
     
    # Compare and print the result
    if(len(s) == 1):
        print('YES')
    else:
        print('NO')
 
# Driver code
if __name__=='__main__':
     
    arr = [ 9, 9, 9, 9, 9, 9, 9 ]
    n = len(arr)
     
    # Function call
    uniqueElement(arr, n)
     
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find if the array
// contains only one distinct element
public static void uniqueElement(int []arr)
{
     
    // Create a set
    HashSet set = new HashSet();
 
    // Traversing the array
    for(int i = 0; i < arr.Length; i++)
    {
        set.Add(arr[i]);
    }
 
    // Compare and print the result
    if (set.Count == 1)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
 
// Driver Code
public static void Main(String []args)
{
    int []arr = { 9, 9, 9, 9, 9, 9, 9 };
 
    // Function call
    uniqueElement(arr);
}
}
 
// This code is contributed by Amit Katiyar

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

高效方法：也可以在不使用任何额外空间的情况下解决此问题。步骤如下：

1. 假定数组的第一个元素是数组中唯一的唯一元素，并将其值存储在变量X中。
2. 然后遍历数组，并检查当前元素是否等于X。
3. 如果发现为真，则继续检查所有数组元素。如果没有发现元素与X不同，则打印“是”。
4. 否则，如果任何数组元素不等于X ，则意味着该数组包含多个唯一元素。因此，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include
using namespace std;
  
// Function to find if the array
// contains only one distinct element
void uniqueElement(int arr[], int n)
{
  // Assume first element to
  // be the unique element
  int x = arr[0];
 
  int flag = 1;
 
  // Traversing the array
  for (int i = 0; i < n; i++)
  {
    // If current element is not
    // equal to X then break the
    // loop and print No
    if (arr[i] != x)
    {
      flag = 0;
      break;
    }
  }
 
  // Compare and print the result
  if (flag == 1)
    cout << "Yes";
  else
    cout << "No";
}
     
// Driver Code
int main()
{
  int arr[] = {9, 9, 9,
               9, 9, 9, 9};
  int n = sizeof(arr) /
          sizeof(arr[0]);
 
  // Function call
  uniqueElement(arr, n);
}
 
// This code is contributed by Chitranayal

Java

// Java program for the above approach
 
import java.util.*;
 
public class Main {
 
    // Function to find if the array
    // contains only one distinct element
    public static void
    uniqueElement(int arr[])
    {
        // Assume first element to
        // be the unique element
        int x = arr[0];
 
        int flag = 1;
 
        // Traversing the array
        for (int i = 0; i < arr.length; i++) {
 
            // If current element is not
            // equal to X then break the
            // loop and print No
            if (arr[i] != x) {
                flag = 0;
                break;
            }
        }
 
        // Compare and print the result
        if (flag == 1)
            System.out.println("Yes");
 
        else
            System.out.println("No");
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int arr[] = { 9, 9, 9, 9, 9, 9, 9 };
 
        // Function call
        uniqueElement(arr);
    }
}

Python3

# Python3 program for the above approach
 
# Function to find if the array
# contains only one distinct element
def uniqueElement(arr):
 
    # Assume first element to
    # be the unique element
    x = arr[0]
 
    flag = 1
 
    # Traversing the array
    for i in range(len(arr)):
 
        # If current element is not
        # equal to X then break the
        # loop and print No
        if(arr[i] != x):
            flag = 0
            break
 
    # Compare and print the result
    if(flag == 1):
        print("Yes")
    else:
        print("No")
 
# Driver Code
 
# Given array arr[]
arr = [ 9, 9, 9, 9, 9, 9, 9 ]
 
# Function call
uniqueElement(arr)
 
# This code is contributed by Shivam Singh

C＃

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find if the array
// contains only one distinct element
public static void uniqueElement(int []arr)
{
     
    // Assume first element to
    // be the unique element
    int x = arr[0];
 
    int flag = 1;
 
    // Traversing the array
    for(int i = 0; i < arr.Length; i++)
    {
         
        // If current element is not
        // equal to X then break the
        // loop and print No
        if (arr[i] != x)
        {
            flag = 0;
            break;
        }
    }
 
    // Compare and print the result
    if (flag == 1)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
 
// Driver code
static public void Main ()
{
    int []arr = { 9, 9, 9, 9, 9, 9, 9 };
 
    // Function call
    uniqueElement(arr);
}
}
 
// This code is contributed by AnkitRai01

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)