# Count the number of elements which are greater than any of element on right side of an array

• Difficulty Level : Basic
• Last Updated : 04 Aug, 2021

Given an array Arr[]. The task is to count the number of elements Arr[i] in the given array such that one or more smaller elements are present on the right side of the element Arr[i] in array.

Examples:

Input: Arr[] = { 3, 9, 4, 6, 7, 5 }
Output: 3
Numbers that counts are: 9, 6, 7
9 – As all numbers are present after 9 are smaller than 9,
6 – 5 is smaller element present after it,
7 – 5 is smaller element which is present after it.

Input: Arr[] = { 3, 2, 1, 2, 3, 4, 5 }
Output:

Approach:
Start traversing array from the last till first element of the array. While traversing maintain a min variable which stores the minimum element till now and a counter variable. Compare min variable with the current element. If min variable is smaller than current element Arr[i], increase the counter and if min is greater than Arr[i] then update the min.

Below is the implementation of the above approach:

## C++

 `// C++ implementation``#include ``using` `namespace` `std;` `// function to return the count``int` `count_greater(``int` `arr[], ``int` `n)``{``    ``int` `min = INT_MAX;``    ``int` `counter = 0;` `    ``// Comparing the given element``    ``// with minimum element till``    ``// occurred till now.``    ``for` `(``int` `i = n - 1; i >= 0; i--) {``        ``if` `(arr[i] > min) {``            ``counter++;``        ``}` `        ``// Updating the min variable``        ``if` `(arr[i] <= min) {``            ``min = arr[i];``        ``}``    ``}` `    ``return` `counter;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 2, 1, 2, 3, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << count_greater(arr, n) << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// function to return the count``static` `int` `count_greater(``int` `arr[], ``int` `n)``{``    ``int` `min = Integer.MAX_VALUE;``    ``int` `counter = ``0``;` `    ``// Comparing the given element``    ``// with minimum element till``    ``// occurred till now.``    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)``    ``{``        ``if` `(arr[i] > min)``        ``{``            ``counter++;``        ``}` `        ``// Updating the min variable``        ``if` `(arr[i] <= min)``        ``{``            ``min = arr[i];``        ``}``    ``}``    ``return` `counter;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``2``, ``1``, ``2``, ``3``, ``4``, ``5` `};``    ``int` `n = arr.length;` `    ``System.out.println(count_greater(arr, n));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation for the above approach``import` `sys` `# function to return the count``def` `count_greater(arr, n) :` `    ``min` `=` `sys.maxsize;``    ``counter ``=` `0``;` `    ``# Comparing the given element``    ``# with minimum element till``    ``# occurred till now.``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``) :``        ``if` `(arr[i] > ``min``) :``            ``counter ``+``=` `1``;``    ` `        ``# Updating the min variable``        ``if` `(arr[i] <``=` `min``) :``            ``min` `=` `arr[i];` `    ``return` `counter;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``3``, ``2``, ``1``, ``2``, ``3``, ``4``, ``5` `];``    ``n ``=` `len``(arr);` `    ``print``(count_greater(arr, n));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// function to return the count``static` `int` `count_greater(``int` `[]arr, ``int` `n)``{``    ``int` `min = ``int``.MaxValue;``    ``int` `counter = 0;` `    ``// Comparing the given element``    ``// with minimum element till``    ``// occurred till now.``    ``for` `(``int` `i = n - 1; i >= 0; i--)``    ``{``        ``if` `(arr[i] > min)``        ``{``            ``counter++;``        ``}` `        ``// Updating the min variable``        ``if` `(arr[i] <= min)``        ``{``            ``min = arr[i];``        ``}``    ``}``    ``return` `counter;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `[]arr = { 3, 2, 1, 2, 3, 4, 5 };``    ``int` `n = arr.Length;``    ` `    ``Console.Write(count_greater(arr, n));``}``}` `// This code is contributed by ajit.`

## Javascript

 ``

Output:

`2`

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

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