大于左边所有元素的数组元素的数量

给定大小为N的数组arr [] ，任务是计算数组元素的数量，以使其左侧的所有元素都严格小于数组元素。
注意：数组的第一个元素将被视为始终满足条件。

例子：

Input: arr[] = { 2, 4, 5, 6 }
Output: 4
Explanation:
Since the array is in increasing order, all the array elements satisfy the condition.
Hence, the count of such elements is equal to the size of the array, i.e. equal to 4.

Input: { 3, 3, 3, 3, 3, 3 }
Output: 1
Explanation: The first array element is the only element satisfying the condition.

方法：
请按照以下步骤解决问题：

• 设置count = 1 ，因为第一个数组元素将被视为满足条件。
• 将第一个数组元素(即arr [0] )设置为maximum 。
• 从i = 1开始遍历数组，并将每个数组元素与当前的最大值进行比较。
• 如果发现数组元素大于当前的最大值，则该元素满足条件，因为其左侧的所有元素都小于它。因此，增加count ，并更新最大值。
• 最后，打印计数。

下面是上述方法的实现：

C++

// C++ program to implement the
//above approach
#include
using namespace std;
  
// Function to return the count
// of array elements with all
// elements to its left smaller
// than it
int count_elements(int arr[], int n)
{
      
    // Stores the count
    int count = 1;
  
    // Stores the maximum
    int max = arr[0];
  
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
          
        // If an element greater
        // than maximum is obtained
        if (arr[i] > max)
        {
              
            // Increase count
            count += 1;
  
            // Update maximum
            max = arr[i];
        }
    }
    return count;
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 1, 4, 6, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
      
    cout << (count_elements(arr, n));
}
  
// This code is contributed by chitranayal

Java

// Java program to implement the
//above approach
import java.util.*;
  
class GFG{
      
// Function to return the count
// of array elements with all
// elements to its left smaller
// than it
static int count_elements(int arr[], int n)
{
      
    // Stores the count
    int count = 1;
  
    // Stores the maximum
    int max = arr[0];
  
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
          
        // If an element greater
        // than maximum is obtained
        if (arr[i] > max)
        {
              
            // Increase count
            count += 1;
  
            // Update maximum
            max = arr[i];
        }
    }
    return count;
}
  
// Driver Code
public static void main(String s[])
{
    int arr[] = { 2, 1, 4, 6, 3 };
    int n = arr.length;
      
    System.out.print(count_elements(arr, n));
} 
}
  
// This code is contributed by rutvik_56

Python3

# Python3 Program to implement
# the above approach
  
# Function to return the count
# of array elements with all
# elements to its left smaller
# than it
  
  
def count_elements(arr):
  
    # Stores the count
    count = 1
  
    # Stores the maximum
    max = arr[0]
  
    # Iterate over the array
    for i in range(1, len(arr)):
  
        # If an element greater
        # than maximum is obtained
        if arr[i] > max:
  
            # Increase count
            count += 1
  
            # Update maximum
            max = arr[i]
    return count
  
  
# Driver Code
arr = [2, 1, 4, 6, 3]
print(count_elements(arr))

C#

// C# program to implement the
// above approach
using System;
  
class GFG{
  
// Function to return the count
// of array elements with all
// elements to its left smaller
// than it
static int count_elements(int[] arr, int n)
{
      
    // Stores the count
    int count = 1;
  
    // Stores the maximum
    int max = arr[0];
  
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
          
        // If an element greater
        // than maximum is obtained
        if (arr[i] > max)
        {
              
            // Increase count
            count += 1;
  
            // Update maximum
            max = arr[i];
        }
    }
    return count;
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 2, 1, 4, 6, 3 };
    int n = arr.Length;
  
    Console.Write(count_elements(arr, n));
}
}
  
// This code is contributed by jrishabh99

输出：

3

时间复杂度： O(N)
辅助空间： O(1)