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:
// 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;
} |
// 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. |
<script> // Javascript implementation // Function to return the count function count_greater(arr, n)
{ let min = Number.MAX_VALUE;
let counter = 0;
// Comparing the given element
// with minimum element till
// occurred till now.
for (let 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 let arr = [ 3, 2, 1, 2, 3, 4, 5 ]; let n = arr.length; document.write(count_greater(arr, n) + "<br>" );
// This code is contributed by rishavmahato348 </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)