Count of array elements which are greater than all elements on its left

Given an array arr[] of size N, the task is to count the number of array elements such that all the elements to its left are strictly smaller than it.
Note: The first element of the array will be considered to be always satisfying the condition.

Examples :

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.

Approach:
Follow the steps below to solve the problem:



  • Set count = 1, as the first array element will be considered to be satisfying the condition.
  • Set the first array element(i.e. arr[0]) as the maximum.
  • Traverse the array starting from i =1, and compare every array element with the current maximum.
  • If an array element is found to be greater than the current maximum, that element satisfies the condition as all the elements on its left are smaller than it. Hence, increase the count, and update the maximum.
  • Finally, print the count.

Below is the implementation of the above approach:

C++

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// C++ program to implement the
//above approach
#include<bits/stdc++.h>
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

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output: 

3

 
 Time Complexity: O(N)
Auxiliary Space: O(1)
 

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