Number of subarrays having sum in a given range

• Difficulty Level : Hard
• Last Updated : 22 Apr, 2021

Given an array arr[] of positive integers and a range (L, R). Find the number of subarrays having sum in the range L to R.
Examples:

Input : arr[] = {1, 4, 6}, L = 3, R = 8
Output : 3
The subarrays are {1, 4}, {4}, {6}.

Input : arr[] = {2, 3, 5, 8}, L = 4, R = 13
Output : 6
The subarrays are {2, 3}, {2, 3, 5}, {3, 5},
{5}, {5, 8}, {8}.

A simple solution is to one by one consider each subarray and find its sum. If the sum lies in the range [L, R], then increment the count. The time complexity of this solution is O(n^2).
An efficient solution is to first find the number of subarrays having sum less than or equal to R. From this count, subtract the number of subarrays having sum less than or equal to L-1. To find the number of subarrays having sum less than or equal to the given value, the linear time method using the sliding window discussed in the following post is used:
Number of subarrays having sum less than or equal to k.
Below is the implementation of above approach:

C++

 // CPP program to find number of subarrays having// sum in the range L to R.#include using namespace std; // Function to find number of subarrays having// sum less than or equal to x.int countSub(int arr[], int n, int x){     // Starting index of sliding window.    int st = 0;     // Ending index of sliding window.    int end = 0;     // Sum of elements currently present    // in sliding window.    int sum = 0;     // To store required number of subarrays.    int cnt = 0;     // Increment ending index of sliding    // window one step at a time.    while (end < n) {         // Update sum of sliding window on        // adding a new element.        sum += arr[end];         // Increment starting index of sliding        // window until sum is greater than x.        while (st <= end && sum > x) {            sum -= arr[st];            st++;        }         // Update count of number of subarrays.        cnt += (end - st + 1);        end++;    }     return cnt;} // Function to find number of subarrays having// sum in the range L to R.int findSubSumLtoR(int arr[], int n, int L, int R){     // Number of subarrays having sum less    // than or equal to R.    int Rcnt = countSub(arr, n, R);     // Number of subarrays having sum less    // than or equal to L-1.    int Lcnt = countSub(arr, n, L - 1);     return Rcnt - Lcnt;} // Driver code.int main(){    int arr[] = { 1, 4, 6 };    int n = sizeof(arr) / sizeof(arr);     int L = 3;    int R = 8;     cout << findSubSumLtoR(arr, n, L, R);    return 0;}

Java

 // Java program to find number// of subarrays having sum in// the range L to R.import java.io.*; class GFG{         // Function to find number    // of subarrays having sum    // less than or equal to x.    static int countSub(int arr[],                        int n, int x)    {             // Starting index of        // sliding window.        int st = 0;             // Ending index of        // sliding window.        int end = 0;             // Sum of elements currently        // present in sliding window.        int sum = 0;             // To store required        // number of subarrays.        int cnt = 0;             // Increment ending index        // of sliding window one        // step at a time.        while (end < n)        {                 // Update sum of sliding            // window on adding a            // new element.            sum += arr[end];                 // Increment starting index            // of sliding window until            // sum is greater than x.            while (st <= end && sum > x)            {                sum -= arr[st];                st++;            }                 // Update count of            // number of subarrays.            cnt += (end - st + 1);            end++;        }             return cnt;    }         // Function to find number    // of subarrays having sum    // in the range L to R.    static int findSubSumLtoR(int arr[], int n,                              int L, int R)    {             // Number of subarrays        // having sum less than        // or equal to R.        int Rcnt = countSub(arr, n, R);             // Number of subarrays        // having sum less than        // or equal to L-1.        int Lcnt = countSub(arr, n, L - 1);             return Rcnt - Lcnt;    }         // Driver code    public static void main (String[] args)    {        int arr[] = { 1, 4, 6 };        int n = arr.length;             int L = 3;        int R = 8;             System.out.println(findSubSumLtoR(arr, n, L, R));    }} // This code is contributed// by Mahadev99

C#

 // C# program to find number// of subarrays having sum in// the range L to R.using System; class GFG{         // Function to find number    // of subarrays having sum    // less than or equal to x.    static int countSub(int[] arr,                        int n, int x)    {             // Starting index of        // sliding window.        int st = 0;             // Ending index of        // sliding window.        int end = 0;             // Sum of elements currently        // present in sliding window.        int sum = 0;             // To store required        // number of subarrays.        int cnt = 0;             // Increment ending index        // of sliding window one        // step at a time.        while (end < n)        {                 // Update sum of sliding            // window on adding a            // new element.            sum += arr[end];                 // Increment starting index            // of sliding window until            // sum is greater than x.            while (st <= end && sum > x)            {                sum -= arr[st];                st++;            }                 // Update count of            // number of subarrays.            cnt += (end - st + 1);            end++;        }             return cnt;    }         // Function to find number    // of subarrays having sum    // in the range L to R.    static int findSubSumLtoR(int[] arr, int n,                              int L, int R)    {             // Number of subarrays        // having sum less than        // or equal to R.        int Rcnt = countSub(arr, n, R);             // Number of subarrays        // having sum less than        // or equal to L-1.        int Lcnt = countSub(arr, n, L - 1);             return Rcnt - Lcnt;    }         // Driver code    public static void Main ()    {        int[] arr = { 1, 4, 6 };        int n = arr.Length;             int L = 3;        int R = 8;             Console.Write(findSubSumLtoR(arr, n, L, R));    }} // This code is contributed// by ChitraNayal

Python 3

 # Python 3 program to find# number of subarrays having# sum in the range L to R. # Function to find number# of subarrays having sum# less than or equal to x.def countSub(arr, n, x):         # Starting index of    # sliding window.    st = 0     # Ending index of    # sliding window.    end = 0     # Sum of elements currently    # present in sliding window.    sum = 0     # To store required    # number of subarrays.    cnt = 0     # Increment ending index    # of sliding window one    # step at a time.    while end < n :                 # Update sum of sliding        # window on adding a        # new element.        sum += arr[end]         # Increment starting index        # of sliding window until        # sum is greater than x.        while (st <= end and sum > x) :            sum -= arr[st]            st += 1         # Update count of        # number of subarrays.        cnt += (end - st + 1)        end += 1     return cnt # Function to find number# of subarrays having sum# in the range L to R.def findSubSumLtoR(arr, n, L, R):         # Number of subarrays    # having sum less    # than or equal to R.    Rcnt = countSub(arr, n, R)     # Number of subarrays    # having sum less than    # or equal to L-1.    Lcnt = countSub(arr, n, L - 1)     return Rcnt - Lcnt # Driver codearr = [ 1, 4, 6 ]n = len(arr)L = 3R = 8print(findSubSumLtoR(arr, n, L, R)) # This code is contributed# by ChitraNayal

PHP

 \$x)        {            \$sum -= \$arr[\$st];            \$st++;        }         // Update count of        // number of subarrays.        \$cnt += (\$end - \$st + 1);        \$end++;    }     return \$cnt;} // Function to find number// of subarrays having sum// in the range L to R.function findSubSumLtoR(&\$arr, \$n, \$L, \$R){    // Number of subarrays    // having sum less    // than or equal to R.    \$Rcnt = countSub(\$arr, \$n, \$R);     // Number of subarrays    // having sum less    // than or equal to L-1.    \$Lcnt = countSub(\$arr, \$n, \$L - 1);     return \$Rcnt - \$Lcnt;} // Driver code.\$arr = array( 1, 4, 6 );\$n = sizeof(\$arr);\$L = 3;\$R = 8;echo findSubSumLtoR(\$arr, \$n, \$L, \$R); // This code is contributed// by ChitraNayal?>

Javascript



Output:

3

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

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