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# Count positions such that all elements before it are greater

• Last Updated : 16 Jun, 2021

Given an array A[], the task is to find the number of positions i in the array such that all elements before A[i] are greater than A[i].
Note: First element is always counted as there is no other element before it.
Examples:

```Input: N = 4, A[] = {2, 1, 3, 5}
Output: 2
The valid positions are 1, 2.

Input : N = 3, A[] = {7, 6, 5}
Output: 3
All three positions are valid positions.```

The idea is to calculate the minimum element every time while traversing the array. That is:

• Initialize the first element as the minimum element.
• Every time a new element arrives, check if this is the new minimum, if so, increment number of valid positions and also initialize minimum to the new minimum.

Below is the implementation of the above approach:

## CPP

 `// C++ Program to count positions such that all``// elements before it are greater` `#include ``using` `namespace` `std;` `// Function to count positions such that all``// elements before it are greater``int` `getPositionCount(``int` `a[], ``int` `n)``{  ``    ``// Count is initially 1 for the first element``    ``int` `count = 1;``    ` `    ``// Initial Minimum``    ``int` `min = a;``    ` `    ``// Traverse the array``    ``for``(``int` `i=1; i

## Java

 `// Java Program to count positions such that all``// elements before it are greater``class` `GFG``{` `// Function to count positions such that all``// elements before it are greater``static` `int` `getPositionCount(``int` `a[], ``int` `n)``{``    ``// Count is initially 1 for the first element``    ``int` `count = ``1``;``    ` `    ``// Initial Minimum``    ``int` `min = a[``0``];``    ` `    ``// Traverse the array``    ``for``(``int` `i = ``1``; i < n; i++)``    ``{``        ``// If current element is new minimum``        ``if``(a[i] <= min)``        ``{``            ``// Update minimum``            ``min = a[i];``            ` `            ``// Increment count``            ``count++;``        ``}``    ``}``    ` `    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``5``, ``4``, ``6``, ``1``, ``3``, ``1` `};``    ``int` `n = a.length;` `    ``System.out.print(getPositionCount(a, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 Program to count positions such that all``# elements before it are greater` `# Function to count positions such that all``# elements before it are greater``def` `getPositionCount(a, n) :` `    ``# Count is initially 1 for the first element``    ``count ``=` `1``;``    ` `    ``# Initial Minimum``    ``min` `=` `a[``0``];``    ` `    ``# Traverse the array``    ``for` `i ``in` `range``(``1``, n) :``    ` `        ``# If current element is new minimum``        ``if``(a[i] <``=` `min``) :``        ` `            ``# Update minimum``            ``min` `=` `a[i];``            ` `            ``# Increment count``            ``count ``+``=` `1``;``    ` `    ``return` `count;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``a ``=` `[ ``5``, ``4``, ``6``, ``1``, ``3``, ``1` `];``    ``n ``=` `len``(a);``    ` `    ``print``(getPositionCount(a, n));``    ` `# This code is contributed by AnkitRai01`

## C#

 `// C# Program to count positions such that all``// elements before it are greater``using` `System;` `class` `GFG``{``    ` `    ``// Function to count positions such that all``    ``// elements before it are greater``    ``static` `int` `getPositionCount(``int` `[]a, ``int` `n)``    ``{``        ``// Count is initially 1 for the first element``        ``int` `count = 1;``        ` `        ``// Initial Minimum``        ``int` `min = a;``        ` `        ``// Traverse the array``        ``for``(``int` `i = 1; i < n; i++)``        ``{``            ``// If current element is new minimum``            ``if``(a[i] <= min)``            ``{``                ``// Update minimum``                ``min = a[i];``                ` `                ``// Increment count``                ``count++;``            ``}``        ``}``        ` `        ``return` `count;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]a = { 5, 4, 6, 1, 3, 1 };``        ``int` `n = a.Length;``    ` `        ``Console.WriteLine(getPositionCount(a, n));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:

`4`

Time Complexity: O(N)

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