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

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:

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// C++ implementation
#include <bits/stdc++.h>
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;
}
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// 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
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# 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
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// 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.
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Output:
2



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