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

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: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 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 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# 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. `

Output:
```2
```

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