# 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

Last Updated : 23 Apr, 2022

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:
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:

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++

 // C++ implementation of the approach #include 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; }

## Java

 // 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

## Python3

 # 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

## C#

 // 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

## PHP

 \$exclud)                 \$count++;         }     }       return \$count; }   // Driver code \$arr = array( 6, 3, 5 );   \$n = count(\$arr) ;   echo countSubarrays(\$arr, \$n);   // This code is contributed by Ryuga ?>

## Javascript



Output:

3

Time Complexity: O(N^2) where N is the length of the array, as are using nested loops to traverse N*N times.

Auxiliary Space: O(N), as we are using pre array of size N while is extra space.

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