# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 remaining array elements.

1. The first loop indicates the beginning of the subarray.
2. The second loop indicate 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:

## CPP

 // C++ implementation of the above approach #include 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 begining 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 begining 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 begining poof a subarray     for i in range(n):            # For loop for ending poof 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 begining 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

Output:

5

Time Complexity: O(N3)

Efficient Approach:

An efficient solution is to use total_sum of given array arr[] that help 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 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 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 begining 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 begining 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 begining 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 begining 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

Output:

5

Time Complexity: O(N2)

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