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++ 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 
// 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 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# 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 
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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