Count the subarray with sum strictly greater than the sum of remaining elements

Given an array arr[] of N positive integers, the task is to count all the subarrays where the sum of subarray elements is strictly greater than the sum of remaining elements.

Examples:

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 6 
Explanation:
Subarrays are: 
{1, 2, 3, 4} – sum of subarray = 10, sum of remaining elements {5} = 5 
{1, 2, 3, 4, 5} – sum of subarray =15, sum of remaining element = 0 
{2, 3, 4} – sum of subarray = 9, sum of remaining elements {1, 5} = 6 
{2, 3, 4, 5} – sum of subarray = 14, sum of remaining elements {1} = 1 
{3, 4, 5} – sum of subarray = 12, sum of remaining elements {1, 2} = 3 
{4, 5} – sum of subarray = 9, sum of remaining elements {1, 2, 3} = 6

Input: arr[] = {10, 9, 12, 6} 
Output: 5 
Explanation:
Sub arrays are : 
{10, 9} – sum of subarray = 19, sum of remaining elements {12, 6} = 18 
{10, 9, 12} – sum of subarray = 31, sum of remaining elements {6} = 6 
{10, 9, 12, 6} – sum of subarray = 37, sum of remaining elements = 0 
{9, 12} – sum of subarray = 21, sum of remaining elements {10, 6} = 16 
{9, 12, 6} – sum of subarray =27, sum of remaining element {10} = 10 

Naive Approach:
A naive approach is to generate the sum of every subarray using three nested loops and check the calculated subarray sum with the sum of the remaining array elements.

1. The first loop indicates the beginning of the subarray.
2. The second loop indicates the ending of the subarray.
3. Inside the second loop, we have for loops to calculate the subarray_sum and remaining array element sum.
4. Increment the counter, when subarray_sum is strictly greater than remaining_sum.

Below is the implementation of the above approach:  

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// sub-arrays with sum strictly greater
// than the remaining elements of array
int Count_subarray(int arr[], int n)
{
    int subarray_sum, remaining_sum, count = 0;

    // For loop for beginning point of a subarray
    for (int i = 0; i < n; i++) {

        // For loop for ending point of the subarray
        for (int j = i; j < n; j++) {

            // Initialise subarray_sum and
            // remaining_sum to 0
            subarray_sum = 0;
            remaining_sum = 0;

            // For loop to calculate
            // the sum of generated subarray
            for (int k = i; k <= j; k++) {
                subarray_sum += arr[k];
            }
            // For loop to calculate the
            // sum remaining array element
            for (int l = 0; l < i; l++) {
                remaining_sum += arr[l];
            }
            for (int l = j + 1; l < n; l++) {
                remaining_sum += arr[l];
            }
            // Checking for condition when
            // subarray sum is strictly greater than
            // remaining sum of array element
            if (subarray_sum > remaining_sum) {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
int main()
{
    int arr[] = { 10, 9, 12, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << Count_subarray(arr, n);
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG
{

// Function to count the number of
// sub-arrays with sum strictly greater
// than the remaining elements of array
static int Count_subarray(int arr[], int n)
{
    int subarray_sum, remaining_sum, count = 0;

    // For loop for beginning point of a subarray
    for (int i = 0; i < n; i++)
    {

        // For loop for ending point of the subarray
        for (int j = i; j < n; j++)
        {

            // Initialise subarray_sum and
            // remaining_sum to 0
            subarray_sum = 0;
            remaining_sum = 0;

            // For loop to calculate
            // the sum of generated subarray
            for (int k = i; k <= j; k++)
            {
                subarray_sum += arr[k];
            }

            // For loop to calculate the
            // sum remaining array element
            for (int l = 0; l < i; l++)
            {
                remaining_sum += arr[l];
            }
            for (int l = j + 1; l < n; l++)
            {
                remaining_sum += arr[l];
            }

            // Checking for condition when
            // subarray sum is strictly greater than
            // remaining sum of array element
            if (subarray_sum > remaining_sum)
            {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 10, 9, 12, 6 };
    int n = arr.length;
    System.out.print(Count_subarray(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python implementation of the above approach

# Function to count the number of
# sub-arrays with sum strictly greater
# than the remaining elements of array
def Count_subarray(arr, n):
    subarray_sum, remaining_sum, count = 0, 0, 0;

    # For loop for beginning point of a subarray
    for i in range(n):

        # For loop for ending point of the subarray
        for j in range(i, n):

            # Initialise subarray_sum and
            # remaining_sum to 0
            subarray_sum = 0;
            remaining_sum = 0;

            # For loop to calculate
            # the sum of generated subarray
            for k in range(i, j + 1):
                subarray_sum += arr[k];


            # For loop to calculate the
            # sum remaining array element
            for l in range(i):
                remaining_sum += arr[l];
            for l in range(j + 1, n):
                remaining_sum += arr[l];

            # Checking for condition when
            # subarray sum is strictly greater than
            # remaining sum of array element
            if (subarray_sum > remaining_sum):
                count += 1;

    return count;

# Driver code
if __name__ == '__main__':
    arr = [ 10, 9, 12, 6];
    n = len(arr);
    print(Count_subarray(arr, n));

# This code is contributed by 29AjayKumar

C#

// C# implementation of the above approach
using System;

class GFG
{

// Function to count the number of
// sub-arrays with sum strictly greater
// than the remaining elements of array
static int Count_subarray(int []arr, int n)
{
    int subarray_sum, remaining_sum, count = 0;

    // For loop for beginning point of a subarray
    for (int i = 0; i < n; i++)
    {

        // For loop for ending point of the subarray
        for (int j = i; j < n; j++)
        {

            // Initialise subarray_sum and
            // remaining_sum to 0
            subarray_sum = 0;
            remaining_sum = 0;

            // For loop to calculate
            // the sum of generated subarray
            for (int k = i; k <= j; k++)
            {
                subarray_sum += arr[k];
            }

            // For loop to calculate the
            // sum remaining array element
            for (int l = 0; l < i; l++)
            {
                remaining_sum += arr[l];
            }
            for (int l = j + 1; l < n; l++)
            {
                remaining_sum += arr[l];
            }

            // Checking for condition when
            // subarray sum is strictly greater than
            // remaining sum of array element
            if (subarray_sum > remaining_sum)
            {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 10, 9, 12, 6 };
    int n = arr.Length;
    Console.Write(Count_subarray(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of
// the above approach

// Function to count the number of
// sub-arrays with sum strictly greater
// than the remaining elements of array
function Count_subarray(arr, n)
{
   var subarray_sum, remaining_sum,
       count = 0;

    var i,j,k,l;
    // For loop for beginning
    // point of a subarray
    for(i = 0; i < n; i++) {

        // For loop for ending point
        // of the subarray
        for (j = i; j < n; j++) {

            // Initialise subarray_sum and
            // remaining_sum to 0
            subarray_sum = 0;
            remaining_sum = 0;

            // For loop to calculate
            // the sum of generated subarray
            for(k = i; k <= j; k++) {
                subarray_sum += arr[k];
            }
            // For loop to calculate the
            // sum remaining array element
            for (l = 0; l < i; l++) {
                remaining_sum += arr[l];
            }
            for (l = j + 1; l < n; l++) {
                remaining_sum += arr[l];
            }
            // Checking for condition when
            // subarray sum is strictly
            // greater than
            // remaining sum of array element
            if (subarray_sum > remaining_sum)
            {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
    var arr =  [10, 9, 12, 6];
    var n = arr.length;
    document.write(Count_subarray(arr, n));

</script>

Output:

5

Time Complexity: O(N³)

Auxiliary Space: O(1)

Efficient Approach:
An efficient solution is to use the total sum of given array arr[] that helps in calculating subarray_sum and remaining_sum. 

1. The total sum of the given array is calculated.
2. Run a for loop where the loop variable i indicate the beginning index of subarray.
3. Another loop, where every j indicate the ending index of the subarray and calculate subarray_sum for every j th index.
4. subarray_sum=arr[i]+arr[i+1]+…..+arr[j] 
   remaining_sum=total_sum – subarray_sum
5. Then, check for condition and increment counter when the subarray sum is strictly greater than the remaining sum of array elements.

Below is the implementation of the above approach.

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

int Count_subarray(int arr[], int n)
{
    int total_sum = 0, subarray_sum,
        remaining_sum, count = 0;

    // Calculating total sum of given array
    for (int i = 0; i < n; i++) {
        total_sum += arr[i];
    }

    // For loop for beginning point of a subarray
    for (int i = 0; i < n; i++) {
        // initialise subarray_sum to 0
        subarray_sum = 0;

        // For loop for calculating
        // subarray_sum and remaining_sum
        for (int j = i; j < n; j++) {

            // Calculating subarray_sum
            // and corresponding remaining_sum
            subarray_sum += arr[j];
            remaining_sum = total_sum - subarray_sum;

            // Checking for the condition when
            // subarray sum is strictly greater than
            // the remaining sum of the array element
            if (subarray_sum > remaining_sum) {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
int main()
{
    int arr[] = { 10, 9, 12, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << Count_subarray(arr, n);
    return 0;
}

Java

// Java implementation of the above approach
class GFG
{

static int Count_subarray(int arr[], int n)
{
    int total_sum = 0, subarray_sum,
        remaining_sum, count = 0;

    // Calculating total sum of given array
    for (int i = 0; i < n; i++)
    {
        total_sum += arr[i];
    }

    // For loop for beginning point of a subarray
    for (int i = 0; i < n; i++)
    {
        // initialise subarray_sum to 0
        subarray_sum = 0;

        // For loop for calculating
        // subarray_sum and remaining_sum
        for (int j = i; j < n; j++)
        {

            // Calculating subarray_sum
            // and corresponding remaining_sum
            subarray_sum += arr[j];
            remaining_sum = total_sum - subarray_sum;

            // Checking for the condition when
            // subarray sum is strictly greater than
            // the remaining sum of the array element
            if (subarray_sum > remaining_sum)
            {
                count += 1;
            }
        }
    }
    return count;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 10, 9, 12, 6 };
    int n = arr.length;
    System.out.print(Count_subarray(arr, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach

def Count_subarray(arr, n) :

    total_sum = 0;
    count = 0;

    # Calculating total sum of given array
    for i in range(n) :
        total_sum += arr[i];

    # For loop for beginning point of a subarray
    for i in range(n) :

        # initialise subarray_sum to 0
        subarray_sum = 0;

        # For loop for calculating
        # subarray_sum and remaining_sum
        for j in range(i, n) :

            # Calculating subarray_sum
            # and corresponding remaining_sum
            subarray_sum += arr[j];
            remaining_sum = total_sum - subarray_sum;

            # Checking for the condition when
            # subarray sum is strictly greater than
            # the remaining sum of the array element
            if (subarray_sum > remaining_sum) :
                count += 1;

    return count;

# Driver code
if __name__ == "__main__" :

    arr = [ 10, 9, 12, 6 ];
    n = len(arr);
    print(Count_subarray(arr, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;

class GFG
{

    static int Count_subarray(int []arr, int n)
    {
        int total_sum = 0, subarray_sum,
            remaining_sum, count = 0;

        // Calculating total sum of given array
        for (int i = 0; i < n; i++)
        {
            total_sum += arr[i];
        }

        // For loop for beginning point of a subarray
        for (int i = 0; i < n; i++)
        {
            // initialise subarray_sum to 0
            subarray_sum = 0;

            // For loop for calculating
            // subarray_sum and remaining_sum
            for (int j = i; j < n; j++)
            {

                // Calculating subarray_sum
                // and corresponding remaining_sum
                subarray_sum += arr[j];
                remaining_sum = total_sum - subarray_sum;

                // Checking for the condition when
                // subarray sum is strictly greater than
                // the remaining sum of the array element
                if (subarray_sum > remaining_sum)
                {
                    count += 1;
                }
            }
        }
        return count;
    }

    // Driver code
    public static void Main()
    {
        int []arr = { 10, 9, 12, 6 };
        int n = arr.Length;
        Console.WriteLine(Count_subarray(arr, n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// javascript implementation of the above approach
function Count_subarray(arr, n)
{
        var total_sum = 0, subarray_sum, remaining_sum, count = 0;

        // Calculating total sum of given array
        for (i = 0; i < n; i++)
        {
            total_sum += arr[i];
        }

        // For loop for beginning point of a subarray
        for (i = 0; i < n; i++)
        {

            // initialise subarray_sum to 0
            subarray_sum = 0;

            // For loop for calculating
            // subarray_sum and remaining_sum
            for (j = i; j < n; j++)
            {

                // Calculating subarray_sum
                // and corresponding remaining_sum
                subarray_sum += arr[j];
                remaining_sum = total_sum - subarray_sum;

                // Checking for the condition when
                // subarray sum is strictly greater than
                // the remaining sum of the array element
                if (subarray_sum > remaining_sum)
                {
                    count += 1;
                }
            }
        }
        return count;
    }

    // Driver code
        var arr = [ 10, 9, 12, 6 ];
        var n = arr.length;
        document.write(Count_subarray(arr, n));

// This code is contributed by todaysgaurav
</script>

Output:

5

Time Complexity: O(N²)

Auxiliary Space: O(1)