Count the number of sub-arrays such that the average of elements present in the sub-array is greater than that not present in the sub-array

Given an array of integers arr[], the task is to count the number of sub-arrays such that the average of elements present in the sub-array is greater than the average of elements that are not present in the sub-array.

Examples:

Input: arr[] = {6, 3, 5}
Output: 3
The sub-arrays are {6}, {5} and {6, 3, 5} because their averages
are greater than {3, 5}, {6, 3} and {} respectively.

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



Approach: The problem can be solved easily by calculating the prefix sum array of the given array. The ith element of the prefix sum array will contain sum of elements up to i. So, the sum of elements between any two indexes i and j can be found using the prefix sum array. Using a nested loop, find all the possible sub-arrays such that its average sum is greater than average of elements not present in the array.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of sub-arrays
// such that the average of elements present
// in the sub-array is greater than the
// average of the elements not present
// in the sub-array
int countSubarrays(int a[], int n)
{
    // Initialize the count variable
    int count = 0;
  
    // Initialize prefix sum array
    int pre[n + 1] = { 0 };
  
    // Preprocessing prefix sum
    for (int i = 1; i < n + 1; i++) {
        pre[i] = pre[i - 1] + a[i - 1];
    }
  
    for (int i = 1; i < n + 1; i++) {
        for (int j = i; j < n + 1; j++) {
  
            // Calculating sum and count
            // to calculate averages
            int sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
            int sum2 = pre[n] - sum1, count2 = ((n - count1) == 0) ? 1 : (n - count1);
  
            // Calculating averages
            int includ = sum1 / count1;
            int exclud = sum2 / count2;
  
            // Increment count if including avg
            // is greater than excluding avg
            if (includ > exclud)
                count++;
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 6, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countSubarrays(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to return the count of sub-arrays
// such that the average of elements present
// in the sub-array is greater than the
// average of the elements not present
// in the sub-array
static int countSubarrays(int a[], int n)
{
    // Initialize the count variable
    int count = 0;
  
    // Initialize prefix sum array
    int []pre = new int[n + 1];
    Arrays.fill(pre, 0);
  
    // Preprocessing prefix sum
    for (int i = 1; i < n + 1; i++)
    {
        pre[i] = pre[i - 1] + a[i - 1];
    }
  
    for (int i = 1; i < n + 1; i++)
    {
        for (int j = i; j < n + 1; j++) 
        {
  
            // Calculating sum and count
            // to calculate averages
            int sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
            int sum2 = pre[n] - sum1, count2 = 
                ((n - count1) == 0) ? 1 : (n - count1);
  
            // Calculating averages
            int includ = sum1 / count1;
            int exclud = sum2 / count2;
  
            // Increment count if including avg
            // is greater than excluding avg
            if (includ > exclud)
                count++;
        }
    }
    return count;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 6, 3, 5 };
    int n = arr.length;
    System.out.println(countSubarrays(arr, n));
}
}
  
// This code is contributed by SURENDRA_GANGWAR

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Python3

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# Python3 implementation of the approach
  
# Function to return the count of sub-arrays
# such that the average of elements present
# in the sub-array is greater than the
# average of the elements not present
# in the sub-array
def countSubarrays(a, n):
      
    # Initialize the count variable
    count = 0
  
    # Initialize prefix sum array
    pre = [0 for i in range(n + 1)]
  
    # Preprocessing prefix sum
    for i in range(1, n + 1):
        pre[i] = pre[i - 1] + a[i - 1]
  
    for i in range(1, n + 1):
        for j in range(i, n + 1):
  
            # Calculating sum and count
            # to calculate averages
            sum1 = pre[j] - pre[i - 1]
            count1 = j - i + 1
            sum2 = pre[n] - sum1
  
            if n-count1 == 0:
                count2 = 1
            else:
                count2 = n - count1
  
            # Calculating averages
            includ = sum1 // count1
            exclud = sum2 // count2
  
            # Increment count if including avg
            # is greater than excluding avg
            if (includ > exclud):
                count += 1
          
    return count
  
# Driver code
arr = [6, 3, 5 ]
n = len(arr)
print(countSubarrays(arr, n))
  
# This code is contributed by mohit kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the count of sub-arrays
// such that the average of elements present
// in the sub-array is greater than the
// average of the elements not present
// in the sub-array
static int countSubarrays(int []a, int n)
{
    // Initialize the count variable
    int count = 0;
  
    // Initialize prefix sum array
    int []pre = new int[n + 1];
    Array.Fill(pre, 0);
  
    // Preprocessing prefix sum
    for (int i = 1; i < n + 1; i++)
    {
        pre[i] = pre[i - 1] + a[i - 1];
    }
  
    for (int i = 1; i < n + 1; i++)
    {
        for (int j = i; j < n + 1; j++) 
        {
  
            // Calculating sum and count
            // to calculate averages
            int sum1 = pre[j] - pre[i - 1], count1 = j - i + 1;
            int sum2 = pre[n] - sum1, count2 = 
                ((n - count1) == 0) ? 1 : (n - count1);
  
            // Calculating averages
            int includ = sum1 / count1;
            int exclud = sum2 / count2;
  
            // Increment count if including avg
            // is greater than excluding avg
            if (includ > exclud)
                count++;
        }
    }
    return count;
}
  
// Driver code
public static void Main()
{
    int []arr = { 6, 3, 5 };
    int n = arr.Length;
    Console.WriteLine(countSubarrays(arr, n));
}
}
  
// This code is contributed by Akanksha Rai

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PHP

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<?php
// PHP implementation of the approach 
  
// Function to return the count of sub-arrays 
// such that the average of elements present 
// in the sub-array is greater than the 
// average of the elements not present 
// in the sub-array 
function countSubarrays($a, $n
    // Initialize the count variable 
    $count = 0; 
  
    // Initialize prefix sum array 
    $pre = array_fill(0, $n + 1, 0); 
  
    // Preprocessing prefix sum 
    for ($i = 1; $i < $n + 1; $i++) 
    
        $pre[$i] = $pre[$i - 1] + $a[$i - 1]; 
    
  
    for ($i = 1; $i < $n + 1; $i++) 
    
        for ($j = $i; $j < $n + 1; $j++) 
        
  
            // Calculating sum and count 
            // to calculate averages 
            $sum1 = $pre[$j] - $pre[$i - 1] ;
            $count1 = $j - $i + 1; 
            $sum2 = $pre[$n] - $sum1
            $count2 = (($n - $count1) == 0) ? 
                   1 : ($n - $count1); 
  
            // Calculating averages 
            $includ = floor($sum1 / $count1); 
            $exclud = floor($sum2 / $count2); 
  
            // Increment count if including avg 
            // is greater than excluding avg 
            if ($includ > $exclud
                $count++; 
        
    
  
    return $count
  
// Driver code 
$arr = array( 6, 3, 5 ); 
  
$n = count($arr) ;
  
echo countSubarrays($arr, $n); 
  
// This code is contributed by Ryuga
?>

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Output:

3

Time Complexity: O(N^2) where N is the length of the array.



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