Number of subarrays having sum less than K

• Difficulty Level : Hard
• Last Updated : 04 May, 2021

Given an array of non-negative numbers and a non-negative 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++

 // CPP program to count// subarrays having sum// less than k.#include 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 Codeint main(){    int array[] = { 1, 11, 2, 3, 15 };    int k = 10;    int size = sizeof(array) / sizeof(array);    int count = countSubarray(array, size, k);    cout << count << "\n";}

Java

 // 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

Python3

 # 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 Codearray = [1, 11, 2, 3, 15]k = 10size = len(array)count = countSubarray(array, size, k);print(count) # This code is contributed by Sam007

C#

 // 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



Javascript


Output :
4

Time complexity : O(n^2).
An efficient solution is based on a sliding window technique that can be used to solve the problem. We use two pointers start and end to represent starting and ending points of the sliding window. (Not that we need to find contiguous parts).
Initially both start and endpoint 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++

 // CPP program to count// subarrays having sum// less than k.#include 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;     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 Codeint main(){    int array[] = { 1, 11, 2, 3, 15 };    int k = 10;    int size = sizeof(array) / sizeof(array);    cout << countSubarrays(array, size, k);}

Java

 // 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;         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

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     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 Codeif __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

C#

 // 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;         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

PHP

 = \$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?>

Javascript


Output:
4

Time complexity : O(n).

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