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Count Subarrays with product of sum and subarray length less than K

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  • Last Updated : 19 Dec, 2022
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Given an array of positive elements arr[] of length N, the task is to count all the subarrays such that the product of the subarray sum and length of the subarray should be strictly less than K

Examples:

Input: arr[] = {6, 2, 1, 4, 3}, K = 10
Output: 6
Explanation: There are six valid subarrays: {6}, {2}, {1}, {4}, {3}, {2, 1}

Input: arr[] = {1, 4, 3}, K = 5
Output: 3

Naive Approach: 

Generate all the subarray and calculate the product of sum and length of the subarray, if it is strictly less than K then increment the count by 1. Finally return the count.

Follow the steps below to implement the above idea:

  • Iterate over all the subarray
    • Calculate the sum of the current subarray and its length
    • Check if the product of sum and length is strictly less than
      • If true, increment the count by 1.
  • Finally, return the count.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

int countSubarrays(vector<int>& arr, int K)
{

    int n = arr.size();
    int count = 0;

    // Iterate over all the subarray
    for (int i = 0; i < n; i++) {
        long long sum = 0;

        for (int j = i; j < n; j++) {

            // Calculate the sum of current subarray and its
            // length
            sum += arr[j];
            int len = j - i + 1;

            // Check if product of sum and length is
            // strictly less than K If true, increment the
            // count by 1.
            if (sum * len < K)
                count++;
        }
    }

    // Finally, return the result.
    return count;
}

// Driver's code
int main()
{
    vector<int> arr = { 6, 2, 1, 4, 3 };
    int K = 10;

    int result = countSubarrays(arr, K);
    cout << result << endl;

    return 0;
}

// This code is contributed by hkdass001

Java

public class Main {
    // Driver's code
    public static void main(String[] args)
    {
        int[] arr = { 6, 2, 1, 4, 3 };
        int K = 10;

        int result = countSubarrays(arr, K);
        System.out.println(result);
    }

    public static int countSubarrays(int[] arr, int K)
    {

        int n = arr.length;
        int count = 0;

        // Iterate over all the subarray
        for (int i = 0; i < n; i++) {
            long sum = 0;

            for (int j = i; j < n; j++) {

                // Calculate the sum of current subarray and
                // its length
                sum += arr[j];
                int len = j - i + 1;

                // Check if product of sum and length is
                // strictly less than K If true, increment
                // the count by 1.
                if (sum * len < K)
                    count++;
            }
        }

        // Finally, return the result.
        return count;
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Python3

def countSubarrays(arr, K):
    n = len(arr)
    count = 0

    # Iterate over all the subarray
    for i in range(n):
        sum = 0

        for j in range(i, n):

            # Calculate the sum of current subarray and its
            # length
            sum += arr[j]
            length = j - i + 1

            # Check if product of sum and length is
            # strictly less than K If true, increment the
            # count by 1.
            if sum * length < K:
                count += 1

    # Finally, return the result.
    return count


# Driver's code
arr = [6, 2, 1, 4, 3]
K = 10

result = countSubarrays(arr, K)
print(result)

# This code is contributed by Tapesh(tapeshdua420)

C#

using System;

class Program {
    // Driver's code
    static void Main(string[] args)
    {
        int[] arr = { 6, 2, 1, 4, 3 };
        int K = 10;

        int result = countSubarrays(arr, K);
        Console.WriteLine(result);
        Console.ReadLine();
    }

    private static int countSubarrays(int[] arr, int K)
    {
        int n = arr.Length;
        int count = 0;

        // Iterate over all the subarray
        for (int i = 0; i < n; i++) {
            long sum = 0;

            for (int j = i; j < n; j++) {

                // Calculate the sum of current subarray and
                // its length
                sum += arr[j];
                int len = j - i + 1;

                // Check if product of sum and length is
                // strictly less than K If true, increment
                // the count by 1.
                if (sum * len < K)
                    count++;
            }
        }

        // Finally, return the result.
        return count;
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Javascript

function countSubarrays(arr, K)
{
    let n = arr.length;
    let count = 0;

    // Iterate over all the subarray
    for (let i = 0; i < n; i++) {
        let sum = 0;

        for (let j = i; j < n; j++) {

            // Calculate the sum of current subarray and its
            // length
            sum += arr[j];
            let len = j - i + 1;

            // Check if product of sum and length is
            // strictly less than K If true, increment the
            // count by 1.
            if (sum * len < K)
                count++;
        }
    }

    // Finally, return the result.
    return count;
}

// Driver's code
let arr = [ 6, 2, 1, 4, 3 ];
let K = 10;

let result = countSubarrays(arr, K);
console.log(result);

// This code is contributed by akashish__
Output

6

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The problem can be solved using Two Pointer or Sliding Window Technique based on the following idea:

Keep two variables start and end, to represent the starting and ending position of any window. 

  • If the current window is a valid subarray then all the possible subarrays that ends at the ending index of the current window is a possible subarray. This number is the same as the length of the current window. 
  • If the current window is not a valid subarray then shrink the window. Keep adding all valid subarray into the result and finally return it.

Follow the steps below to implement the above idea:

  • Initialize start = 0, end = 0, to represent the starting and ending position of the subarray
  • Initialize a variable sum = 0, which represents the sum of elements in the current window.
  • Initialize a variable result = 0 to keep track of the answer.
  • While the end is less than N, do the following
    • Add the element to sum.
    • Calculate the length of the current window
    • Check if sum * length ≥ K i.e.,  if the current window is not valid.
      • Do the following until the current window becomes valid:
        • Remove the contribution of the element at the starting position.
        • Shift the start by 1.
        • Decrement the current window length by 1.
      • Check if the current window is a valid subarray:
        • If true, add (end – start + 1) to result
        • Shift the end by 1
    • Otherwise, Include the (end – start + 1) into the result and shift the end by 1
  • Finally, return the result.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the count of the subarrays
long long countSubarrays(vector<int>& arr, long long k)
{
    long long n = arr.size();

    // Initialise start = 0, end = 0, to
    // represent the starting and ending
    // position of the subarray
    long long start = 0, end = 0;

    // Initialise a variable sum, which
    // represent the sum of elements in the
    // current element.
    long long sum = 0;

    // Initialise a variable result to
    // keep track of answer.
    long long result = 0;

    // While end is less than size of array
    while (end < n) {
        sum += arr[end];
        int len = end - start + 1;

        // Check if current window
        // is not valid
        if (sum * len >= k) {

            // Until current window
            // is not valid
            while (sum * len >= k) {

                // Remove the calculation
                // of element at start
                // position
                sum -= arr[start];

                start++;
                len--;
            }

            // Check if current window
            // is valid subarray
            if (sum * len < k) {

                // If true, add (end - start + 1)
                // into result
                result += end - start + 1;
            }
            end++;
        }

        // Otherwise, Include the (end - start + 1)
        // into result and shift the
        // end by 1
        else {
            result += end - start + 1;
            end++;
        }
    }

    // Finally, return the result
    return result;
}

// Driver code
int main()
{
    vector<int> arr = { 6, 2, 1, 4, 3 };
    int K = 10;

    // Function Call
    cout << countSubarrays(arr, K);

    return 0;
}

// This code is contributed by hkdass001

Java

// Java code to implement the approach

import java.io.*;

class GFG {

    // Function to find the count of the subarrays
    static int countSubarrays(int[] arr, int k)
    {
        int n = arr.length;

        // Initialise start = 0, end = 0, to
        // represent the starting and ending
        // position of the subarray
        int start = 0, end = 0;

        // Initialise a variable sum, which
        // represent the sum of elements in the
        // current element.
        int sum = 0;

        // Initialise a variable result to
        // keep track of answer.
        int result = 0;

        // While end is less than size of array
        while (end < n) {
            sum += arr[end];
            int len = end - start + 1;

            // Check if current window
            // is not valid
            if (sum * len >= k) {

                // Until current window
                // is not valid
                while (sum * len >= k) {

                    // Remove the calculation
                    // of element at start
                    // position
                    sum -= arr[start];

                    start++;
                    len--;
                }

                // Check if current window
                // is valid subarray
                if (sum * len < k) {

                    // If true, add (end - start + 1)
                    // into result
                    result += end - start + 1;
                }
                end++;
            }

            // Otherwise, Include the (end - start + 1)
            // into result and shift the
            // end by 1
            else {
                result += end - start + 1;
                end++;
            }
        }

        // Finally, return the result
        return result;
    }

    public static void main(String[] args)
    {
        int[] arr = { 6, 2, 1, 4, 3 };
        int K = 10;

        // Function call
        System.out.print(countSubarrays(arr, K));
    }
}

// This code is contributed by lokeshmvs21.

Python3

# Python code to implement the approach

# Function to find the count of the subarrays
def countSubarrays(arr, k):
    n = len(arr)

    # Initialise start = 0, end = 0, to
    # represent the starting and ending
    # position of the subarray
    start = 0
    end = 0

    # Initialise a variable sum, which
    # represent the sum of elements in the
    # current element.
    sum = 0

    # Initialise a variable result to
    # keep track of answer.
    result = 0

    # While end is less than size of array
    while end < n:
        sum += arr[end]
        length = end - start + 1

        # Check if current window
        # is not valid
        if sum * length >= k:

            # Until current window
            # is not valid
            while sum * length >= k:

                # Remove the calculation
                # of element at start
                # position
                sum -= arr[start]

                start += 1
                length -= 1

            # Check if current window
            # is valid subarray
            if sum * length < k:

                # If true, add (end - start + 1)
                # into result
                result += end - start + 1

            end += 1

        # Otherwise, Include the (end - start + 1)
        # into result and shift the
        # end by 1
        else:
            result += end - start + 1
            end += 1

    # Finally, return the result
    return result

# Driver code
if __name__ == '__main__':
    arr = [6, 2, 1, 4, 3]
    K = 10

    # Function Call
    print(countSubarrays(arr, K))

# This code is contributed by Tapesh(tapeshdua420)

C#

// C# code to implement the approach
using System;
class GFG {

  // Function to find the count of the subarrays
  static int countSubarrays(int[] arr, int k)
  {
    int n = arr.Length;

    // Initialise start = 0, end = 0, to
    // represent the starting and ending
    // position of the subarray
    int start = 0, end = 0;

    // Initialise a variable sum, which
    // represent the sum of elements in the
    // current element.
    int sum = 0;

    // Initialise a variable result to
    // keep track of answer.
    int result = 0;

    // While end is less than size of array
    while (end < n) {
      sum += arr[end];
      int len = end - start + 1;

      // Check if current window
      // is not valid
      if (sum * len >= k) {

        // Until current window
        // is not valid
        while (sum * len >= k) {

          // Remove the calculation
          // of element at start
          // position
          sum -= arr[start];

          start++;
          len--;
        }

        // Check if current window
        // is valid subarray
        if (sum * len < k) {

          // If true, add (end - start + 1)
          // into result
          result += end - start + 1;
        }
        end++;
      }

      // Otherwise, Include the (end - start + 1)
      // into result and shift the
      // end by 1
      else {
        result += end - start + 1;
        end++;
      }
    }

    // Finally, return the result
    return result;
  }

  public static void Main()
  {
    int[] arr = { 6, 2, 1, 4, 3 };
    int K = 10;

    // Function call
    Console.WriteLine(countSubarrays(arr, K));
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

// Function to find the count of the subarrays
function countSubarrays(arr, k) {
    n = arr.length;

    // Initialise start = 0, end = 0, to
    // represent the starting and ending
    // position of the subarray
    start = 0;
    end = 0;

    // Initialise a variable sum, which
    // represent the sum of elements in the
    // current element.
    sum = 0;

    // Initialise a variable result to
    // keep track of answer.
    result = 0;

    // While end is less than size of array
    while (end < n) {
        sum += arr[end];
        length = end - start + 1;

        // Check if current window
        // is not valid
        if (sum * length >= k) {

            // Until current window
            // is not valid
            while (sum * length >= k) {

                // Remove the calculation
                // of element at start
                // position
                sum -= arr[start];

                start += 1;
                length -= 1;
            }

            // Check if current window
            // is valid subarray
            if (sum * length < k) {

                // If true, add (end - start + 1)
                // into result
                result += end - start + 1;
            }

            end += 1;
        }

        // Otherwise, Include the (end - start + 1)
        // into result and shift the
        // end by 1
        else {
            result += end - start + 1;
            end += 1;
        }
    }

    // Finally, return the result
    return result;
}

// Driver code
arr = [6, 2, 1, 4, 3];
K = 10;

// Function Call
console.log(countSubarrays(arr, K));


// This code is contributed by Tapesh(tapeshdua420)
Output

6

Time Complexity: O(N)
Auxiliary Space: O(1)

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