# 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: 4Explanation:

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: 1Explanation: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*, and**count***update the***maximum.** - Finally, print the
**count.**

Below is the implementation of the above approach:

## C++

`// 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

`// 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` |

## Javascript

`<script>` `// Javascript program to implement the` `// above approach` `// Function to return the count` `// of array elements with all` `// elements to its left smaller` `// than it` `function` `count_elements(arr, n)` `{` ` ` ` ` `// Stores the count` ` ` `let count = 1;` ` ` `// Stores the maximum` ` ` `let max = arr[0];` ` ` `// Iterate over the array` ` ` `for` `(let 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` `let arr = [ 2, 1, 4, 6, 3 ];` `let n = arr.length;` `document.write(count_elements(arr, n));` `// This code is contributed by rishavmahato348` `</script>` |

**Output:**

3

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