Number of subarrays having sum less than K

Given an array of numbers and a number k, find the number of subarrays having sum less than k. We may assume that there is no overflow.

Examples :

Input : arr[] = {2, 5, 6}
        K = 10
Output : 4
The subarrays are {2}, {5}, {6} and
{2, 5},

Input : arr[] = {1, 11, 2, 3, 15}
        K = 10
Output : 4
{1}, {2}, {3} and {2, 3}

A simple solution is to generate all subarrays of the array and then count the number of arrays having sum less than K.
Below is the implementation of above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to count
// subarrays having sum
// less than k.
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number
// of subarrays having sum
// less than k.
int countSubarray(int arr[],
                  int n, int k)
{
    int count = 0;
  
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
  
            // If sum is less than k
            // then update sum and
            // increment count
            if (sum + arr[j] < k) {
                sum = arr[j] + sum;
                count++;
            }
            else {
                break;
            }
        }
    }
  
    return count;
}
  
// Driver Code
int main()
{
    int array[] = { 1, 11, 2, 3, 15 };
    int k = 10;
    int size = sizeof(array) / sizeof(array[0]);
    int count = countSubarray(array, size, k);
    cout << count << "\n";
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count subarrays
// having sum less than k.
import java.io.*;
  
class GFG {
  
    // Function to find number of
    // subarrays having sum less than k.
    static int countSubarray(int arr[],
                             int n, int k)
    {
        int count = 0;
  
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
  
                // If sum is less than
                // k then update sum and
                // increment count
                if (sum + arr[j] < k) {
                    sum = arr[j] + sum;
                    count++;
                }
                else {
                    break;
                }
            }
        }
        return count;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int array[] = { 1, 11, 2, 3, 15 };
        int k = 10;
        int size = array.length;
        int count = countSubarray(array, size, k);
        System.out.println(count);
    }
}
  
// This code is contributed by Sam007

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# python program to count subarrays
# having sum less than k.
  
# Function to find number of subarrays 
# having sum less than k.
def countSubarray(arr, n, k):
    count = 0
  
    for i in range(0, n):
        sum = 0;
        for j in range(i, n):
              
            # If sum is less than k
            # then update sum and 
            # increment count
            if (sum + arr[j] < k):
                sum = arr[j] + sum
                count+= 1
            else:
                break
    return count;
  
  
# Driver Code
array = [1, 11, 2, 3, 15]
k = 10
size = len(array)
count = countSubarray(array, size, k);
print(count)
  
# This code is contributed by Sam007

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to count subarrays
// having sum less than k.
using System;
  
class GFG {
  
    // Function to find number
    // of subarrays having sum
    // less than k.
    static int countSubarray(int[] arr,
                             int n, int k)
    {
        int count = 0;
  
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
  
                // If sum is less than k
                // then update sum and
                // increment count
                if (sum + arr[j] < k) {
                    sum = arr[j] + sum;
                    count++;
                }
                else {
                    break;
                }
            }
        }
        return count;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int[] array = { 1, 11, 2, 3, 15 };
        int k = 10;
        int size = array.Length;
        int count = countSubarray(array, size, k);
        Console.WriteLine(count);
    }
}
  
// This code is contributed by Sam007

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count 
// subarrays having sum 
// less than k.
  
// Function to find number  
// of subarrays having sum 
// less than k.
function countSubarray($arr, $n, $k)
{
    $count = 0;
  
    for ($i = 0; $i < $n; $i++) 
    {
    $sum = 0;
        for ($j = $i; $j < $n; $j++) 
        {
  
            // If sum is less than 
            // k then update sum and 
            // increment count
            if ($sum + $arr[$j] < $k
            {
                $sum = $arr[$j] + $sum;
                $count++;
            }
            else 
            {
                break;
            }
        }
    }
  
    return $count;
}
  
// Driver Code
$array = array(1, 11, 2, 3, 15);
$k = 10;
$size = sizeof($array);
$count = countSubarray($array, $size, $k);
echo $count, "\n";
  
// This code is contributed by ajit
?>

chevron_right


Output :

4


Time complexity :
O(n^2).

An efficient solution is based on sliding window technique that can be used to solve the problem. We use two pointers start and end to represent starting and ending point of sliding window. (Not that we need to find contiguous parts).

Initially both start and end point to the beginning of the array, i.e. index 0. Now, let’s try to add a new element el. There are two possible conditions.

1st case :
If sum is less than k, increment end by one position. So contiguous arrays this step produce are (end – start). We also add el to previous sum. There are as many such arrays as the length of the window.

2nd case :
If sum becomes greater than or equal to k, this means we need to subtract starting element from sum so that the sum again becomes less than k. So we adjust the window’s left border by incrementing start.

We follow the same procedure until end < array size.

Implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to count
// subarrays having sum
// less than k.
#include <bits/stdc++.h>
using namespace std;
  
// Function to find number
// of subarrays having sum
// less than k.
int countSubarrays(int arr[],
                   int n, int k)
{
    int start = 0, end = 0,
        count = 0, sum = arr[0];
  
    while (start < n && end < n) {
  
        // If sum is less than k,
        // move end by one position.
        // Update count and sum
        // accordingly.
        if (sum < k) {
            end++;
  
            if (end >= start)
                count += end - start;
  
            // For last element,
            // end may become n
            if (end < n)
                sum += arr[end];
        }
  
        // If sum is greater than or
        // equal to k, subtract
        // arr[start] from sum and
        // decrease sliding window by
        // moving start by one position
        else {
            sum -= arr[start];
            start++;
        }
    }
  
    return count;
}
  
// Driver Code
int main()
{
    int array[] = { 1, 11, 2, 3, 15 };
    int k = 10;
    int size = sizeof(array) / sizeof(array[0]);
    cout << countSubarrays(array, size, k);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count
// subarrays having sum
// less than k.
import java.io.*;
  
class GFG {
  
    // Function to find number
    // of subarrays having sum
    // less than k.
    static int countSubarray(int arr[],
                             int n, int k)
    {
        int start = 0, end = 0;
        int count = 0, sum = arr[0];
  
        while (start < n && end < n) {
  
            // If sum is less than k,
            // move end by one position.
            // Update count and sum
            // accordingly.
            if (sum < k) {
                end++;
  
                if (end >= start)
                    count += end - start;
  
                // For last element,
                // end may become n.
                if (end < n)
                    sum += arr[end];
            }
  
            // If sum is greater than or
            // equal to k, subtract
            // arr[start] from sum and
            // decrease sliding window by
            // moving start by one position
            else {
                sum -= arr[start];
                start++;
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int array[] = { 1, 11, 2, 3, 15 };
        int k = 10;
        int size = array.length;
        int count = countSubarray(array, size, k);
        System.out.println(count);
    }
}
  
// This code is contributed by Sam007

chevron_right


Python 3

# Python 3 program to count subarrays
# having sum less than k.

# Function to find number of subarrays
# having sum less than k.
def countSubarrays(arr, n, k):

start = 0
end = 0
count = 0
sum = arr[0]

while (start < n and end < n) : # If sum is less than k, move end # by one position. Update count and # sum accordingly. if (sum < k) : end += 1 if (end >= start):
count += end – start

# For last element, end may become n
if (end < n): sum += arr[end] # If sum is greater than or equal to k, # subtract arr[start] from sum and # decrease sliding window by moving # start by one position else : sum -= arr[start] start += 1 return count # Driver Code if __name__ == "__main__": array = [ 1, 11, 2, 3, 15 ] k = 10 size = len(array) print(countSubarrays(array, size, k)) # This code is contributed by ita_c [tabby title="C#"]

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to count
// subarrays having sum
// less than k.
using System;
  
class GFG {
  
    // Function to find number
    // of subarrays having sum
    // less than k.
    static int countSubarray(int[] arr,
                             int n, int k)
    {
        int start = 0, end = 0;
        int count = 0, sum = arr[0];
  
        while (start < n && end < n) {
  
            // If sum is less than k,
            // move end by one position.
            // Update count and sum
            // accordingly.
            if (sum < k) {
                end++;
  
                if (end >= start)
                    count += end - start;
  
                // For last element,
                // end may become n.
                if (end < n)
                    sum += arr[end];
            }
  
            // If sum is greater than or
            // equal to k, subtract
            // arr[start] from sum and
            // decrease sliding window by
            // moving start by one position
            else {
                sum -= arr[start];
                start++;
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int[] array = { 1, 11, 2, 3, 15 };
        int k = 10;
        int size = array.Length;
        int count = countSubarray(array, size, k);
        Console.WriteLine(count);
    }
}
  
// This code is contributed by Sam007

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count 
// subarrays having sum 
// less than k.
  
// Function to find number 
// of subarrays having sum 
// less than k.
function countSubarrays($arr
                       $n, $k)
{
    $start = 0; 
    $end = 0; 
    $count = 0;
    $sum = $arr[0];
  
    while ($start < $n && $end < $n
    {
  
        // If sum is less than k, 
        // move end by one position. 
        // Update count and sum 
        // accordingly.
        if ($sum < $k
        {
            $end++;
  
           if ($end >= $start)
            $count += $end - $start;
  
            // For last element, 
            // end may become n
            if ($end < $n)
            $sum += $arr[$end];
        }
  
        // If sum is greater than or 
        // equal to k, subtract 
        // arr[start] from sum and 
        // decrease sliding window by 
        // moving start by one position
        else
        {
            $sum -= $arr[$start];
            $start++;
        }
    }
  
    return $count;
}
  
    // Driver Code
    $array =array (1, 11, 2, 3, 15);
    $k = 10;
    $size = sizeof($array) ;
    echo countSubarrays($array, $size, $k);
  
// This code is contributed by ajit
?>

chevron_right


Output:

4

Time complexity : O(n).



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Sam007, jit_t, Ita_c



Article Tags :
Practice Tags :


3


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.